Exercises – Short Day

Exercise 1: Diagonalisation of a \(2\times 2\) matrix

We are given the matrix:

\[\begin{split}{\mathbf A}=\left[\begin{array}{cc} 9 & -6\\ 8 & -7 \end{array}\right] \in \mathbb{R}^{2 \times 2}.\end{split}\]

Investigate whether \(\mathbf A\) can be diagonalised, and if so then state an invertible matrix \(\mathbf Q\) and a diagonal matrix \(\mathbf{D}\), such that

\[\mathbf{D}={\mathbf Q}^{-1}\cdot\mathbf A\cdot\mathbf Q.\]

Answer

The matrix \(\mathbf A\) has the two eigenvalues \(-3\) and \(5\), both of which have algebraic and geometric multiplicity \(1\). Hence there exist two linearly independent eigenvectors of \(\mathbf A\), and diagonalisation is therefore possible. Several choices of \(\mathbf D\) and \(\mathbf Q\) are possible; here is one such choice:

\[\begin{split}\mathbf Q =\left[\begin{array}{cc} 1 & 3\\ 2 & 2 \end{array}\right] \,\,\,\,\mathrm{and}\,\,\,\, \mathbf{D}=\left[\begin{array}{cc} -3 & 0\\ 0 & 5 \end{array}\right].\end{split}\]

Exercise 2: Diagonalisation of a linear map

A linear map \(f: \mathbb{R}^3\rightarrow\mathbb{R}^3\) has, with respect to the ordered standard basis \(\varepsilon\) for \(\mathbb{R}^3\), the mapping matrix:

\[\begin{split}{}_\varepsilon [f]_\varepsilon=\left[\begin{array}{ccc} 1 & 0 & 0\\ 1 & 1 & 1\\ 1 & 0 & 2 \end{array}\right] \in \mathbb{R}^{3 \times 3}.\end{split}\]

Question a

Can the map \(f\) be diagonalised?

Hint

The map \(f\) can be diagonalised if and only if its mapping matrix \({}_\varepsilon[f]_\varepsilon\) can be diagonalised. So, first you must find the eigenvalues and eigenspaces of the mapping matrix \({}_\varepsilon[f]_\varepsilon\).

Hint

We find that \({}_\varepsilon[f]_\varepsilon\) has two eigenvalues: one with a geometric multiplicity of \(1\) and another with a geometric multiplicity of \(2\). Based on this, how can we conclude that there exist three linearly independent eigenvectors of \({}_\varepsilon[f]_\varepsilon\) that span \(\mathbb R^3\)?

Answer

\({}_\varepsilon[f]_\varepsilon\) has eigenvalues \(1\) and \(2\) for which \(\mathrm{am}(1)=\mathrm{gm}(1)=2\) and \(\mathrm{am}(2)=\mathrm{gm}(2)=1\). So, the mapping matrix \({}_\varepsilon[f]_\varepsilon\) is diagonalisable, and hence the map \(f\) is also diagonalisable.

In order to find the mentioned geometric multiplicities we must determine the eigenspaces. They are found to be:

\[\begin{split}E_1=\mathrm{Span}_\mathbb{R}\left(\left[\begin{array}{cc} 0 \\ 1 \\ 0 \end{array}\right], \left[\begin{array}{cc} -1 \\ 0 \\ 1 \end{array}\right]\right)\,\,\,\,\mathrm{and}\,\,\,\,\, E_2=\mathrm{Span}_\mathbb{R}\left(\left[\begin{array}{cc} 0 \\ 1 \\ 1 \end{array}\right]\right).\end{split}\]

Question b

Provide an ordered basis \(\beta\) for \(\mathbb{R}^3\) with respect to which the mapping matrix of \(f\) becomes a diagonal matrix. State the change-of-basis matrix \({}_\varepsilon [\mathrm{id}_{\mathbb{R}^3}]_\beta \) that converts \(\beta\)-coordinates into \(\varepsilon\)-coordinates.

Hint

The eigenvectors you found in Question a can be used here.

Answer

The wanted ordered basis \(\beta\) must consist of linearly independent eigenvectors of \(f\). A possible choice is:

\[\begin{split}\beta=\left(\begin{bmatrix}0\\1\\1\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix},\begin{bmatrix}-1\\0\\1\end{bmatrix}\right),\end{split}\]

and the corresponding change-of-basis matrix will then be:

\[\begin{split}{}_\varepsilon[\mathrm{id}_{\mathbb R^3}]_\beta=\begin{bmatrix}0&0&-1\\1&1&0\\1&0&1\end{bmatrix}.\end{split}\]

Question c

State an invertible matrix \( \mathbf Q\) and a diagonal matrix \( \mathbf{D}\) that satisfy

\[\mathbf{D} = {\mathbf Q}^{-1} \cdot {}_\varepsilon [f]_\varepsilon \cdot {\mathbf Q}.\]

Hint

A matrix \(\mathbf Q\) consisting of linearly independent eigenvectors as columns has the wanted properties. Can the change-of-basis matrix from Question b be of use here?

Hint

We have

\[{}_\beta[f]_\beta={}_\beta[\mathrm{id}_{\mathbb R^3}]_\varepsilon\cdot {}_\varepsilon[f]_\varepsilon\cdot {}_\varepsilon[\mathrm{id}_{\mathbb R^3}]_\beta=\left({}_\varepsilon[\mathrm{id}_{\mathbb R^3}]_\beta\right)^{-1}\cdot {}_\varepsilon[f]_\varepsilon\cdot {}_\varepsilon[\mathrm{id}_{\mathbb R^3}]_\beta.\]

Hence, the change-of-basis matrix \({}_\varepsilon[\mathrm{id}_{\mathbb R^3}]_\beta\) has the wanted properties and can be chosen as \(\mathbf Q={}_\varepsilon[\mathrm{id}_{\mathbb R^3}]_\beta\). Note that the columns of \(\mathbf Q\) are linearly independent.

Answer

By choosing

\[\begin{split}\mathbf Q={}_\varepsilon[\mathrm{id}_{\mathbb R^3}]_\beta=\left[\begin{array}{cc} 0 & 0 & -1\\ 1 & 1 & 0\\ 1 & 0 & 1 \end{array}\right]\end{split}\]

we achieve \(\mathbf{D} = {\mathbf Q}^{-1} \cdot {}_\varepsilon [f]_\varepsilon \cdot {\mathbf Q}\), where

\[\begin{split}\mathbf{D}=\left[\begin{array}{cc} 2 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right].\end{split}\]

Exercise 3: Diagonalisation

Question a

We are given the matrix

\[\begin{split}{\mathbf B}=\left[\begin{array}{cc} 3 & 1 & -2 \\ 1 & 3 & 1 \\0 & 0 & 2 \end{array}\right] \in \mathbb{R}^{3 \times 3}.\end{split}\]

Investigate whether \(\mathbf B \) can be diagonalised, and if so then state an invertible matrix \(\mathbf Q\) and a diagonal matrix \(\mathbf{D}\), such that

\[\mathbf{D}={\mathbf Q}^{-1}\cdot\mathbf B\cdot\mathbf Q.\]

Hint

Find all eigenvalues of \(\mathbf B\) and determine their algebraic and geometric multiplicities.

Answer

One eigenvalue of \(\mathbf B\) has an algebraic multiplicity of \(2\) but a geometric multiplicity of \(1\). Hence, matrix \(\mathbf B\) cannot be diagonalised.

Question b

We are given the matrix

\[\begin{split}{\mathbf C}=\left[\begin{array}{cc} 2 & 0 & 0 \\ 1 & 1 & 1 \\-1 & 1 & 1 \end{array}\right] \in \mathbb{R}^{3 \times 3}.\end{split}\]

Investigate whether \(\mathbf C\) can be diagonalised, and if so then state an invertible matrix \(\mathbf Q\) and a diagonal matrix \(\mathbf{D}\), such that

\[\mathbf{D}={\mathbf Q}^{-1}\cdot\mathbf C\cdot\mathbf Q.\]

Answer

Note that \(0\) is one of the eigenvalues.

Diagonalisation is possible since three linearly independent eigenvectors of \(\mathbf C\) exist. There are several correct choices – here is one:

\[\begin{split}\mathbf Q=\left[\begin{array}{cc} 1 & 0 & 1 \\ 1 & -1 & 0 \\ 0 & 1 & -1 \end{array}\right] \,\,\,\,\mathrm{and}\,\,\,\,\mathbf{D}=\left[\begin{array}{cc} 2 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 2 \end{array}\right].\end{split}\]

Exercise 4: Diagonalisation over the complex numbers

We are given the matrix

\[\begin{split}{\mathbf M}=\left[\begin{array}{cc} 3 & 0 & 0 \\ 0 & 1 & -1 \\ 5 & 1 & 1 \end{array}\right] \in \mathbb{C}^{3 \times 3}.\end{split}\]

Question a

Find eigenvalues and associated complex eigenspaces of \(\mathbf M\).

Answer

The eigenvalues are \(1+i\), \(1-i\) and \(3\), each with an algebraic multiplicity of \(1\).

The eigenvectors corresponding to \(\lambda=1+i\) are \(\mathbf v=t_1\cdot \left[\begin{array}{cc} 0 \\ i \\ 1 \end{array}\right]\), where \(t_1\in\mathbb{C}\setminus\{0\}\).

The eigenvectors corresponding to \(\lambda=1-i\) are \(\mathbf v=t_2\cdot \left[\begin{array}{cc} 0 \\ -i \\ 1 \end{array}\right]\), where \(t_2\in\mathbb{C}\setminus\{0\}\).

The eigenvectors corresponding to \(\lambda=3\) are \(\mathbf v=t_3\cdot \left[\begin{array}{cc} 1 \\ -1 \\ 2 \end{array}\right]\), where \(t_3\in\mathbb{C}\setminus\{0\}\).

Question b

Diagonalise \(\mathbf M\), meaning find matrices \(\mathbf Q\) and \(\mathbf{D}\) such that:

\[\mathbf{D}={\mathbf Q}^{-1}\,\mathbf M\,\mathbf Q.\]

Answer

We can choose

\[\begin{split}\mathbf Q=\left[\begin{array}{cc} 0&0&1\\ i&-i&-1\\ 1&1&2 \end{array}\right] \quad\text{and}\quad \mathbf{D}=\left[\begin{array}{cc} 1+i&0&0\\ 0&1-i&0\\ 0&0&3 \end{array}\right],\end{split}\]

which satisfy \(\mathbf{D}={\mathbf Q}^{-1}\,\mathbf M\,\mathbf Q\).

Exercise 5: Diagonalisable or not?

Question a

We are informed that a real matrix \(\mathbf A\in\mathbb R^{2\times 2}\) has two different real eigenvalues. Is matrix \(\mathbf A\) diagonalisable? In other words, is it possible to find a matrix \(\mathbf Q\in\mathbb R^{2\times 2}\) such that \(\mathbf Q^{-1}\cdot \mathbf A\cdot \mathbf Q\) becomes a diagonal matrix?

Answer

Yes.

Question b

We are informed that a real matrix \(\mathbf B\in\mathbb R^{2\times 2}\) has no real eigenvalues. Is matrix \(\mathbf B\) diagonalisable over \(\Bbb R\)?

Answer

No. If there are no (real) eigenvalues then there are also no real eigenvectors in \(\mathbb R^2\). Thus we cannot form a basis for \(\mathbb R^2\) consisting of eigenvectors.

Question c

We are informed that a real matrix \(\mathbf B\in\mathbb R^{2\times 2}\) has no real eigenvalues. Is matrix \(\mathbf B\) diagonalisable over the complex numbers? In other words, is it possible to find a matrix \(\mathbf Q\in\mathbb C^{2\times 2}\) such that \(\mathbf Q^{-1}\cdot \mathbf B\cdot \mathbf Q\) becomes a diagonal matrix?

Hint

Since \(\mathbf B\in\mathbb R^{2\times 2}\), the characteristic polynomial of the matrix has real coefficients. What can then be said about the complex roots of the characteristic polynomial of the matrix? Have a look at Lemma 5.3.3 in the textbook.

Answer

Yes. The characteristic polynomial of the matrix has two different complex roots that are each other’s complex conjugates. Hence we see in a similar way as in Question a that the matrix can be diagonalised over the complex numbers.

Question d

Give an example of a matrix \(\mathbf C\in\mathbb R^{2\times 2}\) that cannot be diagonalised over the real numbers nor over the complex numbers.

Hint

Find a matrix with only one real eigenvalue \(\lambda\) whose geometric multiplicity is \(1\).

Answer

A possible answer is the matrix:

\[\begin{split}\mathbf C=\begin{bmatrix}0&1\\0&0\end{bmatrix}.\end{split}\]

This matrix has the one eigenvalue \(0\) with \(\mathrm{am}(0)=2\) and \(\mathrm{gm}(0)=1\).

Exercise 6: Linear map between polynomial vector spaces

Let \(\mathbb C[Z]\) be the complex vector space consisting of polynomials with complex coefficients. Given a natural number \(n\in\mathbb N\) we define \(V=\mathrm{Span}_{\mathbb C}(1,Z,\ldots,Z^n)\) as the subspace of \(\mathbb C[Z]\) that consists of polynomials of degree no higher than \(n\).

We furthermore define a linear map \(L:V\to V\) by the expression \(L(p(Z))=p'(Z)\), where \(p'(Z)\) denotes the derivative of \(p(Z)\).

Question a

Find the mapping matrix \({}_m[L]_m\), where \(m\) is an ordered basis for \(V\) given by \(m=(1,Z,\ldots,Z^n)\).

Answer

\[\begin{split}{}_m[L]_m=\begin{bmatrix}0&1&0&\cdots&0\\0&0&2&\ddots&\vdots\\\vdots&~&\ddots&\ddots&0\\\vdots&~&~&\ddots&n\\0&~&\cdots&~&0\end{bmatrix}\end{split}\]

Question b

Is \(L\) diagonalisable?

Answer

No.

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