Exercises – Short Day#
Exercise 1: Diagonalization of a \(2\times 2\) Matrix#
We are given the matrix:
Investigate whether \(\mathbf A\) can be diagonalized, and if so then state an invertible matrix \(\mathbf Q\) and a diagonal matrix \(\mathbf{D}\), such that
Answer
The matrix \(\mathbf A\) has the two eigenvalues \(-3\) and \(5\) that both have algebraic and a geometric multiplicities of \(1\). Hence there exist two linearly independent eigenvectors of \(\mathbf A\), and diagonalization is therefore possible. Several choices of \(\mathbf D\) and \(\mathbf Q\) are possible - here is one such choice:
Exercise 2: Diagonalization of a Linear Map#
A linear map \(f: \mathbb{R}^3\rightarrow\ \mathbb{R}^3\) has with respect to the ordered standard basis \(\epsilon\) in \( \mathbb{R}^3 \) the mapping matrix:
Question a#
Can the map \(f\) be diagonalized?
Hint
The map \(f\) can be diagonalized if and only if its mapping matrix \({}_\epsilon[f]_\epsilon\) can be diagonalized. So, first you must find the eigenvalues and eigenspaces of the mapping matrix \({}_\epsilon[f]_\epsilon\).
Hint
It turns out that \({}_\epsilon[f]_\epsilon\) has two eigenvalues: one with a geometric multiplicity of \(1\) and another with a geometric multiplicity of \(2\). How can we based on this conclude that there exist three linearly independent eigenvectors of \({}_\epsilon[f]_\epsilon\) that span \(\mathbb R^3\)?
Answer
\({}_\epsilon[f]_\epsilon\) has the 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 \({}_\epsilon[f]_\epsilon\) is diagonalizable, and hence the map \(f\) is also diagonalizable.
In order to find the mentioned geometric multiplicities we must determine the eigenspaces. They will be found to be:
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 \({}_\epsilon [\mathrm{id}_{\mathbb{R}^3}]_\beta \) that switches from \(\beta\)-coordinates to \(\epsilon\)-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 thus:
and the corresponding change-of-basis matrix will then be:
Question c#
State an invertible matrix \( \mathbf Q\) and a diagonal matrix \( \mathbf{D}\, \) such that
Hint
A matrix \( \mathbf Q\) consisting of linearly independent egenvectors as columns has the wanted properties. Can the change-of-basis matrix from Question b be of use here?
Hint
We have that
Hence, the change-of-basis matrix \({}_\epsilon[\mathrm{id}_{\mathbb R^3}]_\beta\) has the wanted properties and can be chosen as \(\mathbf Q={}_\epsilon[\mathrm{id}_{\mathbb R^3}]_\beta\). Note that the columns of \(\mathbf Q\) are linearly independent.
Answer
By choosing \(\mathbf Q={}_\epsilon[\mathrm{id}_{\mathbb R^3}]_\beta=\left[\begin{array}{cc} 0 & 0 & -1\\ 1 & 1 & 0\\ 1 & 0 & 1 \end{array}\right]\) we achieve that \(\mathbf{D} = {\mathbf Q}^{-1} \cdot {}_e [f]_e \cdot {\mathbf Q}\), where \(\mathbf{D}=\left[\begin{array}{cc} 2 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right]\).
Exercise 3: Diagonalization#
Question a#
We are given the matrix
Investigate whether \(\mathbf B \) can be diagonalized, and if so then state an invertible matrix \(\mathbf Q \) and a diagonal matrix \(\mathbf{D}\,,\) such that
Hint
Find all eigenvalues of \(\mathbf B\) and determine their algebraic and geometric multiplicities.
Answer
On eigenvalue of \(\mathbf B\) has an algebraic multiplicity of \(2\) but a geometric multiplicity of \(1\). Hence, matrix \(\mathbf B\) cannot be diagonalized.
Question b#
We are given the matrix
Investigate whether \(\mathbf C\) can be diagonalized, and if so then state an invertible matrix \(\mathbf Q\) and a diagonal matrix \(\mathbf{D}\,,\) such that
Answer
Note that \(0\) is one of the eigenvalues.
Diagonalization is possible since three linearly independent eigenvectors of \(\mathbf C\) exist. There are several correct choices – here is one:
Exercise 4: Diagonalization over the Complex Numbers#
We are given the matrix
Question a#
Find eigenvalues and associated complex eigenspaces of \(\mathbf M\,.\)
Answer
The eigenvalues are \(1+i\), \(1-i\) and \(3\). They all have 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 \(t_1\in\mathbb{C}\),
the eigenvectors corresponding to \(\lambda=1-i\) are \(\mathbf v=t_2\cdot \left[\begin{array}{cc} 0 \\ -i \\ 1 \end{array}\right]\), where \(t_2\in\mathbb{C}\), and
the eigenvectors corresponding to \(\lambda=3\) are \(\mathbf v=t_3\cdot \left[\begin{array}{cc} 1 \\ -1 \\ 2 \end{array}\right]\), where \(t_3\in\mathbb{C}\).
Question b#
Diagonalize \(\mathbf M\), meaning determine matrices \(\mathbf Q\) and \(\mathbf{D}\) such that:
Answer
If we choose \(\mathbf Q=\left[\begin{array}{cc} 0&0&1\\ i&-i&-1\\ 1&1&2 \end{array}\right] \) and \(\mathbf{D}=\left[\begin{array}{cc} 1+i&0&0\\ 0&1-i&0\\ 0&0&3 \end{array}\right]\) then we achieve as wanted that
Exercise 5: Diagonalizable or not?#
Question a#
We are being informed that a real matrix \(\mathbf A\in\mathbb R^{2\times 2}\) has two different real eigenvalues. Is matrix \(\mathbf A\) diagonalizable? 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 being informed that a real matrix \(\mathbf B\in\mathbb R^{2\times 2}\) has no real eigenvalues. Is matrix \(\mathbf B\) diagonalizable?
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 being informed that a real matrix \(\mathbf B\in\mathbb R^{2\times 2}\) has no real eigenvalues. Is matrix \(\mathbf B\) diagonalizable 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 \(\mathbf B\in\mathbb R^{2\times 2}\), then the characteristic polynomial of the matrix has real coefficient. 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 diagonalized over the complex numbers.
Question d#
Give an example of a matrix \(\mathbf C\in\mathbb R^{2\times 2}\) that cannot be diagonalized over the real numbers and also not over the complex numbers
Answer
We must find a matrix that has only one real eigenvalue \(\lambda\), and this eigenvalue must have a geometric multiplicity of \(1\). A matrix that fulfills this is for example \(\mathbf C=\begin{bmatrix}0&1\\0&0\end{bmatrix}\). This matrix has just one eigenvalue, that being \(0\), and it has \(\mathrm{am}(0)=2\) and \(\mathrm{gm}(0)=1\).
Exercise 6: Linear Map between Polynomial Vector Spaces#
Let \(\mathbb C[Z]\) be a 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 no higher than degree \(n\). 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#
Determine the mapping matrix \({}_m[L]_m\), where \(m\) is an ordered basis for \(V\) given by \(m=(1,Z,\ldots,Z^n).\)
Answer
Question b#
Is \(L\) diagonalizable?
Answer
No.