Diagonalisation

Let $A$ be a square matrix of order $n$ and let $T_A: {\mathbb{F}}^n {\longrightarrow}{\mathbb{F}}^n$ be the corresponding linear transformation. In this section, we ask the question ``does there exist a basis ${\cal B}$ of ${\mathbb{F}}^n$ such that $T_A[{\cal B},{\cal B}],$ the matrix of the linear transformation $T_A,$ is in the simplest possible form."

We know that, the simplest form for a matrix is the identity matrix and the diagonal matrix. In this section, we show that for a certain class of matrices $A,$ we can find a basis ${\cal B}$ such that $T_A[{\cal B},{\cal B}]$ is a diagonal matrix, consisting of the eigenvalues of $A.$ This is equivalent to saying that $A$ is similar to a diagonal matrix. To show the above, we need the following definition.

DEFINITION 6.2.1 (Matrix Diagonalisation)   A matrix $A$ is said to be diagonalisable if there exists a non-singular matrix $P$ such that $P^{-1} A P$ is a diagonal matrix.

Remark 6.2.2   Let $A$ be an $n \times n$ diagonalisable matrix with eigenvalues ${\lambda}_1, {\lambda}_2, \ldots, {\lambda}_n.$ By definition, $A$ is similar to a diagonal matrix $D.$ Observe that $D = {\mbox{diag}}({\lambda}_1, {\lambda}_2, \ldots, {\lambda}_n)$ as similar matrices have the same set of eigenvalues and the eigenvalues of a diagonal matrix are its diagonal entries.

EXAMPLE 6.2.3   Let $A= \left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right].$ Then we have the following:

1. Let $V = {\mathbb{R}}^2.$ Then $A$ has no real eigenvalue (see Example 6.1.8 and hence $A$ doesn't have eigenvectors that are vectors in ${\mathbb{R}}^2.$ Hence, there does not exist any non-singular $2 \times 2$ real matrix $P$ such that $P^{-1} A P$ is a diagonal matrix.
2. In case, $V = {\mathbb{C}}^2 ({\mathbb{C}}),$ the two complex eigenvalues of $A$ are $-i, i$ and the corresponding eigenvectors are $(i, 1)^t$ and $(-i, 1)^t,$ respectively. Also, $(i, 1)^t$ and $(-i, 1)^t$ can be taken as a basis of ${\mathbb{C}}^2 ({\mathbb{C}}).$ Define a $2 \times 2$ complex matrix by $U = \frac{1}{\sqrt{2}}\left[\begin{array}{cc} i & -i \\ 1 & 1 \end{array}\right].$ Then
   $$U^* A U = \left[\begin{array}{cc} -i & 0 \\ 0 & i \end{array}\right].$$

THEOREM 6.2.4   let $A$ be an $n \times n$ matrix. Then $A$ is diagonalisable if and only if $A$ has $n$ linearly independent eigenvectors.

Proof. Let $A$ be diagonalisable. Then there exist matrices $P$ and $D$ such that

$$P^{-1} A P = D = {\mbox{diag}}({\lambda}_1, {\lambda}_2, \ldots, {\lambda}_n).$$

Or equivalently, $A P = P D.$ Let $P = [{\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n].$ Then $A P = P D$ implies that

$$A {\mathbf u}_i = d_i {\mathbf u}_i \;\; {\mbox{ for }} \;\; 1 \leq i \leq n.$$

Since ${\mathbf u}_i$ 's are the columns of a non-singular matrix $P,$ they are non-zero and so for $1 \leq i \leq n,$ we get the eigenpairs $(d_i, {\mathbf u}_i)$ of $A.$ Since, ${\mathbf u}_i$ 's are columns of the non-singular matrix $P,$ using Corollary 4.3.9, we get ${\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n$ are linearly independent.

Thus we have shown that if $A$ is diagonalisable then $A$ has $n$ linearly independent eigenvectors.

Conversely, suppose $A$ has $n$ linearly independent eigenvectors ${\mathbf u}_i, \; 1 \leq i \leq n$ with eigenvalues $\lambda_i.$ Then $A {\mathbf u}_i = \lambda_i {\mathbf u}_i.$ Let $P = [{\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n].$ Since ${\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n$ are linearly independent, by Corollary 4.3.9, $P$ is non-singular. Also,

$$A P = [A {\mathbf u}_1, A {\mathbf u}_2, \ldots, A {\mathbf u}_n] = [{\... ..._1 {\mathbf u}_1, {\lambda}_2 {\mathbf u}_2, \ldots, {\lambda}_n {\mathbf u}_n]$$

$$= [{\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n] \begin{bmat... ...a}_2 & 0 \\ \vdots & \ddots & \vdots \\ 0 & 0 & {\lambda}_n \end{bmatrix}= P D.$$

Therefore the matrix $A$ is diagonalisable. height6pt width 6pt depth 0pt

COROLLARY 6.2.5   let $A$ be an $n \times n$ matrix. Suppose that the eigenvalues of $A$ are distinct. Then $A$ is diagonalisable.

Proof. As $A$ is an $n \times n$ matrix, it has $n$ eigenvalues. Since all the eigenvalues of $A$ are distinct, by Corollary 6.1.17, the $n$ eigenvectors are linearly independent. Hence, by Theorem 6.2.4, $A$ is diagonalisable. height6pt width 6pt depth 0pt

COROLLARY 6.2.6   Let $A$ be an $n \times n$ matrix with ${\lambda}_1, {\lambda}_2, \ldots, {\lambda}_k$ as its distinct eigenvalues and $p({\lambda})$ as its characteristic polynomial. Suppose that for each $i, \; 1 \le i \le k, \; (x - {\lambda}_i)^{m_i}$ divides $p({\lambda})$ but $(x - {\lambda}_i)^{m_i+1}$ does not divides $p({\lambda})$ for some positive integers $m_i$ . Then

$$A \;{\mbox{ is diagonalisable if and only if }} \; \dim\bigl(\ker(A - {\lambda}_i I)\bigr) = m_i \;{\mbox{ for each }} \; i, \; 1 \le i \le k.$$

Or equivalently $A \;{\mbox{ is diagonalisable if and only if }} \; {\mbox{rank}}(A - {\lambda}_i I) = n - m_i \;{\mbox{ for each }} \; i, \; 1 \le i \le k.$

Proof. As $A$ is diagonalisable, by Theorem 6.2.4, $A$ has $n$ linearly independent eigenvalues. Also, $\sum\limits_{i=1}^k m_i = n$ as $\deg( p({\lambda})) = n$ . Hence, for each eigenvalue ${\lambda}_i, \; 1 \le i \le k$ , $A$ has exactly $m_i$ linearly independent eigenvectors. Thus, for each $i, \; 1 \le i \le k$ , the homogeneous linear system $(A - {\lambda}_i I) {\mathbf x}= {\mathbf 0}$ has exactly $m_i$ linearly independent vectors in its solution set. Therefore, $\dim\bigl(\ker(A - {\lambda}_i I)\bigr) \ge m_i$ . Indeed $\dim\bigl(\ker(A - {\lambda}_i I)\bigr) = m_i$ for $1 \le i \le k$ follows from a simple counting argument.

Now suppose that for each $i, \; 1 \le i \le k, \;\dim\bigl(\ker(A - {\lambda}_i I)\bigr) = m_i$ . Then for each $i, \; 1 \le i \le k$ , we can choose $m_i$ linearly independent eigenvectors. Also by Corollary 6.1.17, the eigenvectors corresponding to distinct eigenvalues are linearly independent. Hence $A$ has $n = \sum\limits_{i=1}^k m_i$ linearly independent eigenvectors. Hence by Theorem 6.2.4, $A$ is diagonalisable. height6pt width 6pt depth 0pt

EXAMPLE 6.2.7  

1. Let $A=\left[\begin{array}{ccc}2 & 1 & 1\\ 1 & 2 & 1\\ 0 & -1 & 1 \end{array}\right].$ Then $\det ( A - {\lambda}I) = (2 - {\lambda})^2 (1 - {\lambda}).$ Hence, $A$ has eigenvalues $1, 2, 2.$ It is easily seen that $\bigl(1, (1,0, -1)^t \bigr)$ and $(\bigl( 2, (1,1,-1)^t \bigr)$ are the only eigenpairs. That is, the matrix $A$ has exactly one eigenvector corresponding to the repeated eigenvalue $2.$ Hence, by Theorem 6.2.4, the matrix $A$ is not diagonalisable.
2. Let $A=\left[\begin{array}{ccc}2 & 1 & 1\\ 1 & 2 & 1\\ 1 & 1 & 2 \end{array}\right].$ Then $\det ( A - {\lambda}I) = (4 - {\lambda})(1 - {\lambda})^2.$ Hence, $A$ has eigenvalues $1, 1, 4.$ It can be easily verified that $(1,-1,0)^t$ and $(1,0,-1)^t$ correspond to the eigenvalue $1$ and $(1,1,1)^t$ corresponds to the eigenvalue $\; 4.$ Note that the set $\{ (1, -1, 0)^t, (1, 0, -1)^t \}$ consisting of eigenvectors corresponding to the eigenvalue $1$ are not orthogonal. This set can be replaced by the orthogonal set $\{(1,0,-1)^t, (1,-2,1)^t\}$ which still consists of eigenvectors corresponding to the eigenvalue $1$ as $(1, -2, 1) = 2 (1,-1,0) - (1,0,-1)$ . Also, the set $\{(1,1,1), (1,0,-1), (1,-2,1)\}$ forms a basis of ${\mathbb{R}}^3.$ So, by Theorem 6.2.4, the matrix $A$ is diagonalisable. Also, if $U = \left[\begin{array}{ccc} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1... ...frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \end{array}\right]$ is the corresponding unitary matrix then $U^* A U = {\mbox{diag}}(4,1,1).$

   Observe that the matrix $A$ is a symmetric matrix. In this case, the eigenvectors are mutually orthogonal. In general, for any $n \times n$ real symmetric matrix $A,$ there always exist $n$ eigenvectors and they are mutually orthogonal. This result will be proved later.

EXERCISE 6.2.8  

1. By finding the eigenvalues of the following matrices, justify whether or not $A = P D P^{-1}$ for some real non-singular matrix $P$ and a real diagonal matrix $D.$
   $i) \;\; \begin{bmatrix}\cos \theta & \sin \theta \\ - \sin \theta & \cos \thet... ...{bmatrix}\cos \theta & \sin \theta \\ \sin \theta & - \cos \theta \end{bmatrix}$ for any $\theta$ with $0 \leq \theta \leq 2 \pi.$
2. Are the two matrices $\begin{bmatrix}2 & 1 \\ -1 & 0 \end{bmatrix}$ and $\begin{bmatrix}2 & i \\ i & 0 \end{bmatrix}$ diagonalisable?
3. Find the eigenvalues and eigenvectors of $A = [a_{ij}]_{n \times n}$ , where $a_{ij} = a$ if $i=j$ and $b$ otherwise.
4. Let $A$ be an $n \times n$ matrix and $B$ an $m \times m$ matrix. Suppose $C = \begin{bmatrix}A & {\mathbf 0}\\ {\mathbf 0}& B \end{bmatrix}.$ Then show that $C$ is diagonalisable if and only if both $A$ and $B$ are diagonalisable.
5. Let $T : {\mathbb{R}}^5 \longrightarrow {\mathbb{R}}^5$ be a linear transformation with ${\mbox{rank }}(T - I) = 3$ and
   $${\cal N}(T) = \{ (x_1, x_2, x_3, x_4, x_5) \in {\mathbb{R}}^5 \mid x_1 + x_4 + x_5 = 0, \; x_2 + x_3 = 0 \}.$$
   Then
   1. determine the eigenvalues of $T?$
   2. find the number of linearly independent eigenvectors corresponding to each eigenvalue?
   3. is $T$ diagonalisable? Justify your answer.
6. Let $A$ be a non-zero square matrix such that $A^2 = {\mathbf 0}.$ Show that $A$ cannot be diagonalised. [Hint: Use Remark 6.2.2.]
7. Are the following matrices diagonalisable?
   $i) \;\;\begin{bmatrix}1 & 3 & 2 & 1 \\ 0 & 2 & 3 & 1 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 4 \end{bmatrix}, \;\;\;\;$ $ii) \;\; \begin{bmatrix}1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix},\;\;\;\;$ $iii) \;\;\begin{bmatrix}1 & -3 & 3 \\ 0 & -5 & 6 \\ 0 & -3 & 4 \end{bmatrix}.$