🔬 Tutorial problems eta

🔬 Tutorial problems eta#

\(\eta\).1#

Consider an \((n \times n)\) Vandermonde matrix [this one can be named :)] of the form

\[\begin{split} V = \begin{bmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \end{bmatrix} \end{split}\]

Show that the determinant of \(V\) is given by

\[ \det(V) = \Pi_{j<i \leqslant n}(x_i-x_j) \]

for the cases \(n=2\), \(n=3\) and \(n=4\)

Properties of the determinants help in finding an elegant solution.

Using the fact that the determinant is invariant to Gaussian elementary row operations we perform the following derivation:

\[\begin{split} \det \begin{pmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \end{pmatrix} = \det \begin{pmatrix} 1 & 0 & \cdots & 0 \\ x_1 & x_2-x_1 & \cdots & x_n-x_1 \\ x_1^2 & x_2^2-x_1^2 & \cdots & x_n^2-x_1^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^{n-1} & x_2^{n-1}-x_1^{n-1} & \cdots & x_n^{n-1}-x_1^{n-1} \end{pmatrix} = \end{split}\]

(expand along the first row)

\[\begin{split} = \det \begin{pmatrix} x_2-x_1 & x_3-x_1 & \cdots & x_n-x_1 \\ x_2^2-x_1^2 & x_3^2-x_1^2 & \cdots & x_n^2-x_1^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_2^{n-1}-x_1^{n-1} & x_3^{n-1}-x_1^{n-1} & \cdots & x_n^{n-1}-x_1^{n-1} \end{pmatrix} = \end{split}\]

(take out common factor from each row)

\[\begin{split} = (x_2-x_1)(x_3-x_1)\dots(x_n-x_1) \det \begin{pmatrix} 1 & 1 & \cdots & 1 \\ \frac{x_2^2-x_1^2}{x_2-x_1} & \frac{x_3^2-x_1^2}{x_3-x_1} & \cdots & \frac{x_n^2-x_1^2}{x_n-x_1} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{x_2^{n-1}-x_1^{n-1}}{x_2-x_1} & \frac{x_3^{n-1}-x_1^{n-1}}{x_3-x_1} & \cdots & \frac{x_n^{n-1}-x_1^{n-1}}{x_n-x_1} \\ \end{pmatrix} \end{split}\]

It is not hard to show (by long division of polynomials and mathematical induction) that

\[ \frac{a^{k}-b^{k}}{a-b} =a^{k-1}+a^{k-2}b + \dots + ab^{k-2} + b^{k-1} =\sum_{i=1}^{k} a^{k-i}b^{i-1} \]

Continuing the derivation:

\[\begin{split} = \prod_{i=2}^n (x_i-x_1) \det \begin{pmatrix} 1 & \cdots & 1 \\ x_2+x_1 & \cdots & x_n+x_1 \\ \vdots & \ddots & \vdots \\ \sum_{i=1}^{n-1} x_2^{n-1-i}x_1^{i-1} & \cdots & \sum_{i=1}^{n-1} x_n^{n-1-i}x_1^{i-1} \end{pmatrix} = \end{split}\]

We can then repeat the same procedure:

  • subtract the first row from all other

  • notice regularities in the polynomials

  • expand the determinant along the first row

  • take out common factors from each row

It is then obvious how to prove the main formula by mathematical induction.

\(n=2\)

\[\begin{split} \det \begin{pmatrix} 1 & 1 \\ x & y \end{pmatrix} =y-x \end{split}\]

\(n=3\)

\[\begin{split} \det \begin{pmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{pmatrix} = \det \begin{pmatrix} 1 & 0 & 0 \\ x & y-x & z-x \\ x^2 & y^2-x^2 & z^2-x^2 \end{pmatrix} = \end{split}\]
\[\begin{split} = \det \begin{pmatrix} y-x & z-x \\ y^2-x^2 & z^2-x^2 \end{pmatrix} = \end{split}\]
\[\begin{split} = (y-x)(z-x) \det \begin{pmatrix} 1 & 1 \\ x+y & x+z \end{pmatrix} = \end{split}\]
\[ = (y-x)(z-x)(x+z-x-y)=(y-x)(z-x)(z-y) \]

\(n=4\)

Follow the same scheme, using

  • \(x^2-y^2=(x-y)(x+y)\)

  • \(x^3-y^3=(x-y)(x^2+xy+y^2)\)

\(\eta\).2#

For each of the linear maps defined by the following matrices

\[\begin{split} T_1 = \begin{pmatrix} 4/3 & -2/3 & 0 \\ -1/3 & 5/3 & 0 \\ 0 & 0 & -1 \end{pmatrix} \end{split}\]
\[\begin{split} T_2 = \begin{pmatrix} 4 & 0 & 1 \\ -2 & 1 & 0 \\ -2 & 0 & 1 \end{pmatrix} \end{split}\]
\[\begin{split} T_3 = \begin{pmatrix} 5 & 0 & 1 \\ 1 & 1 & 0 \\ -7 & 1 & 0 \end{pmatrix} \end{split}\]
\[\begin{split} T_4 = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & -1 \\ 0 & -1 & 3 \end{pmatrix} \end{split}\]

perform the following tasks:

  1. Find eigenvalues

  2. Find at least one eigenvector for each eigenvalue

  3. Form a new basis from the eigenvectors (normalized or not)

  4. Compute the transformation matrix to the new basis

  5. Find the matrix \(T\) in the new basis and verify that it is diagonal

See example in the lecture notes

\(T_1\)

1.

To find eigenvalues solve

\[\begin{split} \det \begin{pmatrix} 4/3-\lambda & -2/3 & 0 \\ -1/3 & 5/3-\lambda & 0 \\ 0 & 0 & -1-\lambda \end{pmatrix} = -(1+\lambda) \begin{pmatrix} 4/3-\lambda & -2/3 \\ -1/3 & 5/3-\lambda \end{pmatrix} = \end{split}\]
\[ = -(1+\lambda) \left[\left(\frac{4}{3}-\lambda \right)\left(\frac{5}{3}-\lambda \right)-\frac{2}{3\cdot3}\right] = -(1+\lambda) (\lambda^2-3\lambda+\frac{4\cdot 5}{9}-\frac{2}{9}) = \]
\[ = -(1+\lambda)(\lambda^2-3\lambda+2) = -(1+\lambda)(\lambda-1)(\lambda-2) \]

Therefore the eigenvalues are \(\lambda_1=-1\), \(\lambda_2=1\) and \(\lambda_3=2\)

2.

To find eigenvectors plug the eigenvalues one by one to \(T_1-\lambda I\) and investigate the implications of the resulting system of equations. We should expect to find multiple eigenvectors for each eigenvalue, and therefore are looking for a formula rather than a usual answer.

\[\begin{split} T_1 - \lambda_1 I=0 \iff \begin{pmatrix} 7/3 & -2/3 & 0 \\ -1/3 & 8/3 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{split}\]

Thus, any value of \(z\) will do, for example \(z=1\). To find \(x\) and \(y\) we can use the first two equations (multiplied by 3 right away):

\[\begin{split} \left\{ \begin{array}{rcl} 7x - 2y &=& 0 \\ -x + 8y &=& 0 \end{array} \right. \implies \left\{ \begin{array}{l} x = 0 \\ y = 0 \end{array} \right. \end{split}\]

Therefore, the vector \(v_1 = (0,0,1)\) is an eigenvector for \(\lambda_1=-1\).

\[\begin{split} T_1 - \lambda_2 I=0 \iff \begin{pmatrix} 1/3 & -2/3 & 0 \\ -1/3 & 2/3 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \iff \end{split}\]
\[\begin{split} \iff \begin{pmatrix} 1 & -2 & 0 \\ 1 & -2 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{split}\]

Obviously, all vectors of the form \((2a,a,0)\) are eigenvectors for \(\lambda_2=1\). In particular, \(v_2 = (2/\sqrt{5},1/\sqrt{5},0)\) has a norm of 1.

\[\begin{split} T_1 - \lambda_3 I=0 \iff \begin{pmatrix} -2/3 & -2/3 & 0 \\ -1/3 & -1/3 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \iff \end{split}\]
\[\begin{split} \iff \begin{pmatrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{split}\]

Now, all vectors of the form \((a,-a,0)\) are eigenvectors for \(\lambda_3=2\). In particular, \(v_3 = (1/\sqrt{2},-1/\sqrt{2},0)\) has a norm of 1.

3.

We have chosen the eigenvectors in such a way that they are already normalized, i.e. have length of 1. To verify, observe

\[\begin{split} \begin{array}{l} \|v_1\| = \sqrt{0^2+0^2+1^2} = 1 \\ \|v_2\| = \sqrt{(2/\sqrt{5})^2+(1/\sqrt{5})^2+0^2} = 1 \\ \|v_2\| = \sqrt{(1/\sqrt{2})^2+(-1/\sqrt{2})^2+0^2} = 1 \end{array} \end{split}\]

It’s easy to verify that vectors \(v_1, v_2, v_3\) are linearly independent, and therefore the set

\[\begin{split} \{v_1, v_2, v_3\} = \left\{ \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 2/\sqrt{5} \\ 1/\sqrt{5} \\ 0 \end{pmatrix}, \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} \right\} \end{split}\]

forms a normalized basis in \(\mathbb{R}^3\)

4.

The transformation matrix is a matrix with columns formed from the vectors of the new basis expressed in coordinates (``the language’’) of the old basis.

\[\begin{split} P = \begin{pmatrix} 0 & 2/\sqrt{5} & 1/\sqrt{2} \\ 0 & 1/\sqrt{5} & -1/\sqrt{2} \\ 1 & 0 & 0 \end{pmatrix} \end{split}\]

5.

The matrix \(T\) and the matrix \(T'\) in the new basis are related as

\[ T = P T' P^{-1} \quad \iff \quad T' = P^{-1} T P \]

In any case, we need \(P^{-1}\). In order to find the inverse of the \(P\) matrix, we make the following argument. By definition, \(PP^{-1} = I\), and therefore the columns of the unknown matrix \(P^{-1}\) (denoted below \(p'_i\), \(i=1,2,3\)) are solutions of the following three systems of equations:

\[\begin{split} P p'_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \quad P p'_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \quad P p'_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \end{split}\]

We can find the solutions of all three systems by Gaussian elimination performing elementary row operations on an ``extra’’ augmented matrix with three columns in place of the right-hand side.

\[\begin{split} \left(\begin{array}{ccc|ccc} 0 & 2/\sqrt{5} & 1/\sqrt{2} & 1 & 0 & 0 \\ 0 & 1/\sqrt{5} & -1/\sqrt{2} & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 \end{array}\right) \longrightarrow \end{split}\]
\[\begin{split} \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & -\sqrt{5}/\sqrt{2} & 0 & \sqrt{5} & 0 \\ 0 & 2/\sqrt{5} & 1/\sqrt{2} & 1 & 0 & 0 \end{array}\right) \longrightarrow \end{split}\]
\[ \frac{1}{\sqrt{2}} + \left(-\frac{\sqrt{5}}{\sqrt{2}}\right)\left(-\frac{2}{\sqrt{5}}\right) = \frac{1}{\sqrt{2}} + \frac{2}{\sqrt{2}} = \frac{3}{\sqrt{2}} \]
\[\begin{split} \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & -\sqrt{5}/\sqrt{2} & 0 & \sqrt{5} & 0 \\ 0 & 0 & 3/\sqrt{2} & 1 & -2 & 0 \end{array}\right) \longrightarrow \end{split}\]
\[\begin{split} \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & -\sqrt{5}/\sqrt{2} & 0 & \sqrt{5} & 0 \\ 0 & 0 & 1 & \sqrt{2}/3 & -2\sqrt{2}/3 & 0 \end{array}\right) \longrightarrow \end{split}\]
\[ \sqrt{5} + \frac{\sqrt{5}}{\sqrt{2}} \left(-\frac{2 \sqrt{2}}{3} \right) = \frac{\sqrt{5}}{3} \]
\[\begin{split} \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & \sqrt{5}/3 & \sqrt{5}/3 & 0 \\ 0 & 0 & 1 & \sqrt{2}/3 & -2\sqrt{2}/3 & 0 \end{array}\right) \longrightarrow \end{split}\]

Therefore, the inverse of the \(P\) matrix is given by

\[\begin{split} P^{-1} = \begin{pmatrix} 0 & 0 & 1 \\ \sqrt{5}/3 & \sqrt{5}/3 & 0 \\ \sqrt{2}/3 & -2\sqrt{2}/3 & 0 \end{pmatrix} \end{split}\]

Additional exercise: verify that \(PP^{-1} = I\).

Now we can compute \(P^{-1} T_1 P\):

\[\begin{split} \begin{pmatrix} 0 & 0 & 1 \\ \sqrt{5}/3 & \sqrt{5}/3 & 0 \\ \sqrt{2}/3 & -2\sqrt{2}/3 & 0 \end{pmatrix} \begin{pmatrix} 4/3 & -2/3 & 0 \\ -1/3 & 5/3 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & 2/\sqrt{5} & 1/\sqrt{2} \\ 0 & 1/\sqrt{5} & -1/\sqrt{2} \\ 1 & 0 & 0 \end{pmatrix} = \end{split}\]
\[\begin{split} \begin{pmatrix} 0 & 0 & -1 \\ \frac{4\sqrt{5}}{9}-\frac{\sqrt{5}}{9} & \frac{-2\sqrt{5}}{9}+\frac{5\sqrt{5}}{9} & 0 \\ \frac{4\sqrt{2}}{9}+\frac{2\sqrt{2}}{9} & \frac{-2\sqrt{2}}{9}-\frac{10\sqrt{2}}{9} & 0 \end{pmatrix} \begin{pmatrix} 0 & 2/\sqrt{5} & 1/\sqrt{2} \\ 0 & 1/\sqrt{5} & -1/\sqrt{2} \\ 1 & 0 & 0 \end{pmatrix} = \end{split}\]
\[\begin{split} \begin{pmatrix} 0 & 0 & -1 \\ \frac{\sqrt{5}}{3} & \frac{\sqrt{5}}{3} & 0 \\ \frac{2\sqrt{2}}{3} & \frac{-4\sqrt{2}}{3} & 0 \end{pmatrix} \begin{pmatrix} 0 & 2/\sqrt{5} & 1/\sqrt{2} \\ 0 & 1/\sqrt{5} & -1/\sqrt{2} \\ 1 & 0 & 0 \end{pmatrix} = \end{split}\]
\[\begin{split} \begin{pmatrix} -1 & 0 & 0 \\ 0 & \frac{2}{3} + \frac{1}{3} & \frac{\sqrt{5}}{3\sqrt{2}} - \frac{\sqrt{5}}{3\sqrt{2}} \\ 0 & \frac{4\sqrt{2}}{3\sqrt{5}} - \frac{-4\sqrt{2}}{3\sqrt{5}} & \frac{2}{3} + \frac{4}{3} \end{pmatrix} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix} \end{split}\]

\(P^{-1} T_1 P\) is diagonal with eigenvalues on the main diagonal!

\(T_2\)

1.

To find eigenvalues solve

\[\begin{split} \det \begin{pmatrix} 4-\lambda & 0 & 1 \\ -2 & 1-\lambda & 0 \\ -2 & 0 & 1-\lambda \end{pmatrix} = \end{split}\]

(expanding along the top row)

\[\begin{split} =(4-\lambda)(1-\lambda)^1 + 2(1-\lambda) = \\ (1-\lambda)(\lambda^2-5\lambda+6) = \\ (1-\lambda)(\lambda-2)(\lambda-3) \end{split}\]

Therefore the eigenvalues are \(\lambda_1=1\), \(\lambda_2=2\) and \(\lambda_3=3\)

2.

To find eigenvectors plug the eigenvalues one by one to \(T_2-\lambda I\) and investigate the implications of the resulting system of equations. We should expect to find multiple eigenvectors for each eigenvalue, and therefore are looking for a formula rather than a usual answer.

\[\begin{split} T_2 - \lambda_1 I=0 \iff \begin{pmatrix} 3 & 0 & 1 \\ -2 & 0 & 0 \\ -2 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{split}\]

Doing Gaussian-Jordan elimination we have

\[\begin{split} \left(\begin{array}{ccc|c} 3 & 0 & 1 & 0 \\ -2 & 0 & 0 & 0 \\ -2 & 0 & 0 & 0 \end{array}\right) \rightarrow \left(\begin{array}{ccc|c} 1 & 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{2}{3} & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \rightarrow \end{split}\]
\[\begin{split} \left(\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \iff \left\{ \begin{array}{l} x = 0 \\ y \text{ is free} \\ z = 0 \end{array} \right. \end{split}\]

In other words, eigenvector \(v_1\) is of the form \((0,p,0)\) where \(p\in\mathbb{R}\). Let us this time not impose the normalization and take \(v_1 = (0,1,0)\) as the first eigenvector.

\[\begin{split} T_2 - \lambda_2 I=0 \iff \begin{pmatrix} 2 & 0 & 1 \\ -2 & -1 & 0 \\ -2 & 0 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{split}\]
\[\begin{split} \left(\begin{array}{ccc|c} 2 & 0 & 1 & 0 \\ -2 & -1 & 0 & 0 \\ -2 & 0 & -1 & 0 \end{array}\right) \rightarrow \left(\begin{array}{ccc|c} 1 & 0 & \frac{1}{2} & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \rightarrow \end{split}\]
\[\begin{split} \left(\begin{array}{ccc|c} 1 & 0 & \frac{1}{2} & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \iff \left\{ \begin{array}{l} x = -\frac{1}{2} z \\ y =z \\ z \text{ is free} \end{array} \right. \end{split}\]

In other words, eigenvector \(v_2\) is of the form \((-\frac{1}{2}p,p,p)\) where \(p\in\mathbb{R}\). Again, let us not impose the normalization and take \(v_2 = (-1,2,2)\) as the second eigenvector.

\[\begin{split} T_2 - \lambda_3 I=0 \iff \begin{pmatrix} 1 & 0 & 1 \\ -2 & -2 & 0 \\ -2 & 0 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{split}\]
\[\begin{split} \left(\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ -2 & -2 & 0 & 0 \\ -2 & 0 & -2 & 0 \end{array}\right) \rightarrow \left(\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & -2 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \rightarrow \end{split}\]
\[\begin{split} \left(\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \iff \left\{ \begin{array}{l} x = -z \\ y =z \\ z \text{ is free} \end{array} \right. \end{split}\]

In other words, eigenvector \(v_3\) is of the form \( (-p,p,p)\) where \(p\in\mathbb{R}\). Let us take \(v_3 = (-\frac{1}{2},1,1)\) as the second eigenvector.

3.

We have chosen the eigenvectors in a way that they are not normalized, and let’s try to see if this approach results in a diagonal matrix \(T'\) in the new basis anyway.

\[\begin{split} \{v_1, v_2, v_3\} = \left\{ \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix}, \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \right\} \end{split}\]

forms a basis in \(\mathbb{R}^3\)

4.

The transformation matrix is a matrix with columns formed from the vectors of the new basis expressed in coordinates (``the language’’) of the old basis.

\[\begin{split} P = \begin{pmatrix} 0 & -1 & -1 \\ 1 & 2 & 1 \\ 0 & 2 & 1 \end{pmatrix} \end{split}\]

5.

Again, the matrix \(T\) and the matrix \(T'\) in the new basis are related as

\[ T = P T' P^{-1} \quad \iff \quad T' = P^{-1} T P \]

We find \(P^{-1}\) in the same way by going Gaussian-Jordan elimination

\[\begin{split} \left(\begin{array}{ccc|ccc} 0 & -1 & -1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 & 1 & 0 \\ 0 & 2 & 1 & 0 & 0 & 1 \end{array}\right) \longrightarrow \end{split}\]
\[\begin{split} \left(\begin{array}{ccc|ccc} 1 & 2 & 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 2 & 0 & 1 \end{array}\right) \longrightarrow \end{split}\]
\[\begin{split} \left(\begin{array}{ccc|ccc} 1 & 0 & -1 & 2 & 1 & 0 \\ 0 & 1 & 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & -2 & 0 & -1 \end{array}\right) \longrightarrow \end{split}\]
\[\begin{split} \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -2 & 0 & -1 \end{array}\right) \longrightarrow \end{split}\]

Therefore, the inverse of the \(P\) matrix is given by

\[\begin{split} P^{-1} = \begin{pmatrix} 0 & 1 & -1 \\ 1 & 0 & 1 \\ -2 & 0 & -1 \end{pmatrix} \end{split}\]

Additional exercise: verify that \(PP^{-1} = I\).

Now we can compute \(P^{-1} T_2 P\):

\[\begin{split} \begin{pmatrix} 0 & 1 & -1 \\ 1 & 0 & 1 \\ -2 & 0 & -1 \end{pmatrix} \begin{pmatrix} 4 & 0 & 1 \\ -2 & 1 & 0 \\ -2 & 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & -1 & -1 \\ 1 & 2 & 1 \\ 0 & 2 & 1 \end{pmatrix} = \end{split}\]
\[\begin{split} \begin{pmatrix} 0 & 1 & -1 \\ 2 & 0 & 2 \\ -6 & 0 & -3 \end{pmatrix} \begin{pmatrix} 0 & -1 & -1 \\ 1 & 2 & 1 \\ 0 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \end{split}\]

\(P^{-1} T_2 P\) is diagonal with eigenvalues on the main diagonal!

\(T_3\)

1.

To find eigenvalues solve

\[\begin{split} \det \begin{pmatrix} 5-\lambda & 0 & 1 \\ 1 & 1-\lambda & 0 \\ -7 & 1 & -\lambda \end{pmatrix} = (5-\lambda)(1-\lambda)(-\lambda)+1+7(1-\lambda) = \\ =(1-\lambda)(\lambda^2-5\lambda+7) + 1 = \lambda^2-5\lambda+7 - \lambda^3+5\lambda^2-7\lambda +1 =\\ =-\lambda^3+6\lambda^2-12\lambda+8 = -(\lambda-2)^3 \end{split}\]

Therefore the only eigenvalue is \(\lambda=2\), this root is repeated three times.

2.

To find eigenvectors plug the eigenvalues one by one to \(T_3-\lambda I\) and investigate the implications of the resulting system of equations.

Because the eigenvalue is repeated, we should expect difficulties finding enough eigenvectors to form a new basis — we need at least three linearly independent eigenvectors in \(\mathbb{R}^3\)!

\[\begin{split} T_3 - \lambda I=0 \iff \begin{pmatrix} 3 & 0 & 1 \\ 1 & -1 & 0 \\ -7 & 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{split}\]

Doing Gaussian-Jordan elimination we have

\[\begin{split} \left(\begin{array}{ccc|c} 3 & 0 & 1 & 0 \\ 1 & -1 & 0 & 0 \\ -7 & 1 & -2 & 0 \end{array}\right) \rightarrow \left(\begin{array}{ccc|c} 1 & 0 & \frac{1}{3} & 0 \\ 0 & -1 & -\frac{1}{3} & 0 \\ 0 & 1 & \frac{1}{3} & 0 \end{array}\right) \rightarrow \end{split}\]
\[\begin{split} \left(\begin{array}{ccc|c} 1 & 0 & \frac{1}{3} & 0 \\ 0 & 1 & \frac{1}{3} & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \iff \left\{ \begin{array}{l} x = -\frac{1}{3}z \\ y = - \frac{1}{3}z \\ z = \text{ is free} \end{array} \right. \end{split}\]

In other words, all eigenvectors have the form \((-\frac{1}{3}p,-\frac{1}{3}p,p)\) where \(p\in\mathbb{R}\).

In order to form a basis from eigenvectors, we need three linearly independent of them, which is impossible in this case because there is only one free parameter! In other words, all eigenvectors we can come up with will lie within the same line (one degree of freedom), and thus we can not even have two linearly inpendent eigenvectors, let alone three.

3. - 5.

\(T_3\) is not diagonalizable through eigendecomposition.

\(T_4\)

1.

First of all, note that \(T_4\) is symmetric with real entries, therefore eigenvalue decomposition definitely be possible – unlike the last case. To find eigenvalues solve

\[\begin{split} \det \begin{pmatrix} 2-\lambda & 0 & 0 \\ 0 & 3-\lambda & -1 \\ 0 & -1 & 3-\lambda \end{pmatrix} = (2-\lambda)\big((3-\lambda)^2-1\big) = \\ =(2-\lambda)(9-6\lambda+\lambda^2-1) = (2-\lambda)(\lambda^2-6\lambda+8) = \\ =(2-\lambda)(\lambda -2)(\lambda-4) = -(\lambda-2)^2(\lambda-4) \end{split}\]

Therefore eigenvalues are \(\lambda_1=4\) and a repeated one \(\lambda_{2,3}=2\). Yet, again, we should be able to diagonalize \(T_4\) because it is symmetric!

2.

To find eigenvectors plug the eigenvalues one by one to \(T_4-\lambda I\) and investigate the implications of the resulting system of equations.

Because the eigenvalue is repeated, we should expect to do more work than usual, but the basis using eigenvectors should still be possible to find.

\[\begin{split} T_4 - \lambda_1 I=0 \iff \begin{pmatrix} -2 & 0 & 0 \\ 0 & -1 & -1 \\ 0 & -1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{split}\]
\[\begin{split} \left(\begin{array}{ccc|c} -2 & 0 & 0 & 0 \\ 0 & -1 & -1 & 0 \\ 0 & -1 & -1 & 0 \end{array}\right) \rightarrow \left(\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \rightarrow \iff \left\{ \begin{array}{l} x = 0 \\ y = -z \\ z = \text{ is free} \end{array} \right. \end{split}\]

In other words, the eigenvectors corresponding to \(\lambda_1\) have the form \((0,-p,p)\) where \(p\in\mathbb{R}\).

\[\begin{split} T_4 - \lambda_{2,3} I=0 \iff \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & -1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{split}\]

It is clear immediately, that the only restriction placed by this linear system of equations is that \(y=z\). In other words, there are no restrictions on two of the three variables, and we can express the eigenvectors corresponding to \(\lambda_{2,3}\) as \((p,q,q)\) where \(p, q\in\mathbb{R}\) are parameters.

3.

In order to form a basis from eigenvectors, we need three linearly independent of them. Fortunately, there is enough degrees of freedom in the parameters (one from the first eigenvalue and two from the second) to have three linearly independent eigenvectors. For example, \((0,-1,1)\), \((1,0,0)\) and \((0,1,1)\)

4.

The transformation matrix is a matrix with columns formed from the vectors of the new basis expressed in coordinates (``the language’’) of the old basis.

\[\begin{split} P = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 1 \\ 1 & 0 & 1 \end{pmatrix} \end{split}\]

5.

Again, the matrix \(T\) and the matrix \(T'\) in the new basis are related as

\[ T = P T' P^{-1} \quad \iff \quad T' = P^{-1} T P \]

We find \(P^{-1}\) in the same way by going Gaussian-Jordan elimination

\[\begin{split} \left(\begin{array}{ccc|ccc} 0 & 1 & 0 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 & 1 \end{array}\right) \longrightarrow \end{split}\]
\[\begin{split} \left(\begin{array}{ccc|ccc} 1 & 0 & 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 & 1 & 1 \end{array}\right) \longrightarrow \end{split}\]
\[\begin{split} \left(\begin{array}{ccc|ccc} 1 & 0 & 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & \frac{1}{2} & \frac{1}{2} \end{array}\right) \longrightarrow \end{split}\]
\[\begin{split} \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & -\frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & \frac{1}{2} & \frac{1}{2} \end{array}\right) \longrightarrow \end{split}\]
\[\begin{split} P^{-1} = \begin{pmatrix} 0 & -\frac{1}{2} & \frac{1}{2} \\ 1 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} \end{pmatrix} \end{split}\]

Additional exercise: verify that \(PP^{-1} = I\).

Now we can compute \(P^{-1} T_2 P\):

\[\begin{split} \begin{pmatrix} 0 & -\frac{1}{2} & \frac{1}{2} \\ 1 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & -1 \\ 0 & -1 & 3 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 1 \\ 1 & 0 & 1 \end{pmatrix} = \end{split}\]
\[\begin{split} \begin{pmatrix} 0 & -\frac{1}{2} & \frac{1}{2} \\ 1 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} 0 & 2 & 0 \\ -4 & 0 & 2 \\ 4 & 0 & 2 \end{pmatrix} = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \end{split}\]

We see again that \(P^{-1} T_4 P\) is diagonal with eigenvalues on the main diagonal, even though one of the eigenvalues is repeated twice.

\(\eta\).3#

Derive canonical equations for the three conic sections in Euclidean space by using the properties stated in the definitions for:

  • ellipse

  • parabola

  • hyperbola

For example, for ellipse, assume that the focal points are located at \((c,0)\) and \((-c,0)\), and consider an arbitrary point \((x,y) \in \mathbb{R}^2\) which satisfies the defining property of the ellipse. Then, simplify the expression to arrive at the canonical equation of the ellipse, and give interpretations for the parameters \((a,b)\).

Perform additional task for each curve:

  1. Ellipse: derive expressions for the coordinates of the focal points using standard parameters \(a\) and \(b\)

  2. Parabola: explain the geometric meaning of parameter \(p\)

  3. Hyperbola: verify what change of variables leads to the equation \(xy=1\) for the same curve

Start with the definitions of the conic section, and formulate the defining geometric properties of each curve.

Example of what needs to be done, for a circle. Using the definition that a circle is the set of points equidistant from a fixed point (center), we derive the canonical equation of the circle \(x^2 + y^2 = r^2\).

Let a center be situated in the origin. Then, for an arbitrary point \((x,y) \in \mathbb{R}^2\), the distance from the center is given by the Euclidean norm \(\sqrt{x^2 + y^2}\). The defining property of the circle is that this distance is constant, and equal to the radius \(r\). Therefore, the canonical equation of the circle is

\[ \sqrt{x^2 + y^2} = r \quad \iff \quad x^2 + y^2 = r^2 \]

Ellipse

Let focal points of ellipse be located at \((c,0)\) and \((-c,0)\), and consider an arbitrary point \((x,y) \in \mathbb{R}^2\) which satisfies the defining property of the ellipse, namely the sum of distances from \((x,y)\) to the focal points is constant denoted \(2d\).

\[\begin{split} \begin{array}{rcl} \sqrt{(x-c)^2 + y^2} + \sqrt{(x+c)^2 + y^2} &=& 2d \\ \sqrt{(x-c)^2 + y^2} &=& 2d - \sqrt{(x+c)^2 + y^2} \\ (x-c)^2 + y^2 &=& 4d^2 - 4d\sqrt{(x+c)^2 + y^2} + (x+c)^2 + y^2 \\ 4d\sqrt{(x+c)^2 + y^2} &=& 4d^2 +4xc \\ d^2(x+c)^2 + d^2y^2 &=& d^4 +2xcd^2 +x^2c^2 \\ x^2d^2 + 2xcd^2 + c^2d^2 + d^2y^2 &=& d^4 +2xcd^2 +x^2c^2 \\ x^2 (d^2-c^2) + y^2d^2 &=& d^4 - c^2d^2 \end{array} \end{split}\]
\[ \frac{x^2}{d^2} + \frac{y^2}{d^2-c^2} = 1 \]

That is, the canonical parameters \(a^2 = d^2\) and \(b^2 = d^2-c^2\). Ellipse intersects the \(x\)-axis at \(x = \pm a\) and \(y\)-axis at \(y = \pm b\).

The focal points in canonical parameters are located at \((\sqrt{a^2-b^2},0)\) and \((-\sqrt{a^2-b^2},0)\).

Parabola

Let the focal point of parabola be located at \((0,c)\) and the dirextrix at \(y=-c\). Consider an arbitrary point \((x,y) \in \mathbb{R}^2\) which satisfies the defining property of the parabola, namely the distance from \((x,y)\) to the focal point is equal to the distance from \((x,y)\) to the directrix \(y=-c\). The latter is equal to the distance from the projection of the point \((x,y)\) to point \((0,-c)\) where the directrix crosses \(y\)-axis.

\[\begin{split} \begin{array}{rcl} \sqrt{x^2+(y-c)^2} &=& y+c \\ x^2+(y-c)^2 &=& y^2+2yc+c^2 \\ x^2+y^2-2yc+c^2 &=& y^2+2yc+c^2 \\ x^2 &=& 4yc \end{array} \end{split}\]

That is, the canonical parameter for parabola is \(p = 2c\), the distance from the focal point to the directrix. The vertex of the parabola is in the origin, half way between.

Hyperbola

Again, let the focal points be located at \((c,0)\) and \((-c,0)\), and consider an arbitrary point \((x,y) \in \mathbb{R}^2\) which satisfies the defining property of the hyperbola, namely the difference of distances from \((x,y)\) to the focal points is constant denoted \(2d\).

\[\begin{split} \begin{array}{rcl} \sqrt{(x+c)^2 + y^2} - \sqrt{(x-c)^2 + y^2} &=& 2d \\ \sqrt{(x+c)^2 + y^2} &=& 2d + \sqrt{(x-c)^2 + y^2} \\ (x+c)^2 + y^2 &=& 4d^2 + 4d\sqrt{(x-c)^2 + y^2} + (x-c)^2 + y^2 \\ 4d\sqrt{(x-c)^2 + y^2} &=& 4xc - 4d^2 \\ d^2(x-c)^2 + d^2y^2 &=& x^2c^2-2xcd^2+d^4 \\ x^2d^2 -2xcd^2 +c^2d^2 + d^2y^2 &=& x^2c^2-2xcd^2+d^4 \\ x^2 (c^2-d^2) - y^2d^2 &=& d^2(c^2 - d^2) \end{array} \end{split}\]
\[ \frac{x^2}{d^2} - \frac{y^2}{d^2-c^2} = 1 \]

That is, again the canonical parameters \(a^2 = d^2\) and \(b^2 = d^2-c^2\).

\(\eta\).4#

Consider a quadratic form

\[ Q(x) = x_1^2 + x_2^2 + x_3^2 + 4x_4^2 + 2x_1x_2 - 2x_1x_3 + 4x_1x_4 + 6x_2x_4 - 4x_3x_4 \]

Convert this quadratic form to the canonical form this time by algebraic derivations and not eigenvalue decomposition.

Recall all the involved definitions, in particular that a canonical form of quadratic form is just a linear combination of squares of variables.

The algebraic derivation should aim at completing squares, with the appropriate change of bases to effectively change the variables. For example, an expression \(x^2 + 2xy + y^2\) is a square of a variable \(z = x+y\), which can be attained after the change of bases with the transition matrix

\[\begin{split} T = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \\ T \colon \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} x \\ x+y \end{pmatrix} \end{split}\]

First, note that completing the square with many variables is not too hard:

\[\begin{split} \begin{array}{l} (a+b)^2 = a^2 + 2ab + b^2 \\ (a+b+c)^2 = a^2 + 2ab + b^2 + 2c(a+b) + c^2 \\ (a+b+c+d)^2 = a^2 + b^2 + c^2 + d^2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd\\ \left(\sum_{i=1}^N x_i\right)^2 = \sum_{i=1}^N x_i^2 + 2\sum_{i=1}^N\sum_{j=i+1}^N x_i x_j \end{array} \end{split}\]

(It’s a quadratic form with the matrix which elements are all 1!)

Now, let’s try to complete the square for the given quadratic form:

\[\begin{split} \begin{array}{rl} Q(x) =& x_1^2 + x_2^2 + x_3^2 + 4x_4^2 + 2x_1x_2 - 2x_1x_3 + 4x_1x_4 + 6x_2x_4 - 4x_3x_4 =\\=& \big[x^2_1 + 2x_1x_2 - 2x_1x_3 + 4x_1x_4 \big] + x_2^2 + x_3^2 + 4x_4^2 + 6x_2x_4 - 4x_3x_4 =\\=& \big(x_1 + x_2 - x_3 + 2x_4 \big)^2 + 2x_2x_3 -4x_2x_4 + 4x_3x_4 + 6x_2x_4 - 4x_3x_4 =\\=& \big(x_1 + x_2 - x_3 + 2x_4 \big)^2 +2x_2x_3 + 2x_2x_4 \end{array} \end{split}\]

Introduce the first change of variables:

\[\begin{split} \begin{array}{l} y_1 = x_1 + x_2 - x_3 + 2x_4 \\ y_i = x_i, i \in \{2,3,4\} \end{array} \quad \Rightarrow \quad y = \begin{pmatrix} 1 & 1 & -1 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} x \end{split}\]

After the change of variables from \(x\) to \(y\) the quadratic form becomes

\[\begin{split} \begin{array}{rl} Q(y) =& y_1^2 +2y_2y_3 + 2y_2y_4 =\\=& y_1^2 + [y^2_2 + y^2_3 + y^2_4 + 2y_2y_3 + 2y_2y_4 + 2y_3y_4 ] - y^2_2 - y^2_3 - y^2_4 - 2y_3y_4 =\\=& y_1^2 + (y_2 + y_3 + y_4)^2 - y^2_2 - y^2_3 - y^2_4 - 2y_3y_4 =\\=& y_1^2 + (y_2 + y_3 + y_4)^2 - y^2_2 - [y^2_3 + y^2_4 + 2y_3y_4] =\\=& y_1^2 + (y_2 + y_3 + y_4)^2 - y^2_2 - (y_3 + y_4)^2 \end{array} \end{split}\]

Introduce the second change of variables:

\[\begin{split} \begin{array}{l} t_1 = y_1 \\ t_2 = y_2 \\ t_3 = y_2 + y_3 + y_4 \\ t_4 = y_3 + y_4 \end{array} \quad \Rightarrow \quad t = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{pmatrix} y \end{split}\]

After the change of variables from \(x\) to \(y\) to \(t\) the quadratic takes the canonical form

\[ Q(t) = t_1^2 - t_2^2 + t_3^2 - t_4^2 \]

The complete change of variables from \(x\) to \(t\) can be thought of as a composite linear map

\[\begin{split} t = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & -1 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} x = \begin{pmatrix} 1 & 1 & -1 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{pmatrix} x \end{split}\]

Quick check reveals that the determinant of this matrix is zero, and therefore, its inverse does not exists, but we can still use it to transform the variables to express the original quadratic form in the canonical form.

In matrix notation, the original quadratic form is written as

\[\begin{split} Q(x) = x^T A x, \quad A = \begin{pmatrix} 1 & 1 & -1 & 2 \\ 1 & 1 & 0 & 3 \\ -1 & 0 & 1 & -2 \\ 2 & 3 & -2 & 4 \end{pmatrix} \end{split}\]

In the diagonalized (canonical) form, it is

\[\begin{split} Q(t) = t^T T^T \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} T t \end{split}\]
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