A square matrix $A$ is diagonalisable if there is a non-singular matrix $P$ such that $P^{-1}AP$ is a diagonal matrix. Here we consider the question: given a matrix, is it diagonalisable? If so, how do we find $P$?

The secret to constructing such a $P$ is to let the columns of $P$ be the eigenvectors of $A$. We immediately have that $AP=PD,$ where $D$ is a diagonal matrix with eigenvalues on the diagonal. We know from section 15.1 that $P$ is invertible if and only if the columns of $P$ are linearly independent.

Hence, we have the following result:

The $n\times n$ matrix $A$ is diagonalisable if and only if $A$ has $n$ linearly independent eigenvectors.

Is the matrix

diagonalisable?

We know this matrix has eigenvalues $-3, -1, -4,$ with corresponding eigenvectors

We form the matrix

whose columns are these eigenvectors.

We can check directly that $P^{-1}AP=D,$ where $$\left(\begin{array}{rrr}-3&0&0\\0&-1&0\\0&0&-4\end{array}\right)=D$$ is a diagonal matrix with the eigenvalues on the diagonal.

So steps for diagonalising an $n\times n$ matrix $A$:

1. Find eigenvalues $\lambda_1, \lambda_2, \cdots,$ and the corresponding eigenvectors $\v_1, \v_2, \cdots,$ of $A.$
2. Check if $A$ has $n$ linearly independent eigenvectors.
3. If no, $A$ is not diagonalizable.
   If yes, $A$ is diagonalizable. In this case, form matrix $P=(\v_1|\v_2|\cdots|\v_n)$. Then $P^{-1}$ exists and $P$ diagonalizes $A,$ i.e., $P^{-1}AP=diag\{\lambda_1, \cdots, \lambda_n\}.$

Two matrices $A$ and $B$ are similar if there is a non-singular matrix $P$ such that $B=P^{-1}AP.$

The statements "$A$ is diagonalisable" and "$A$ is similar to a diagonal matrix" are equivalent.

16.1.1 Theorem (similar matrices)

Similar matrices have the same eigenvalues.

16.1.1 Theorem (similar matrices)

Similar matrices have the same eigenvalues.

In fact, if $B=P^{-1}AP$ and $\v$ is an eigenvector of $A$ corresponding to eigenvalue $\lambda,$ then $P^{-1}\v$ is an eigenvector of $B$ corresponding to eigenvalue $\lambda$. This is because

Let the matrix $A$ be $n\times n$ with $n$ linearly independent eigenvectors $\v_1,\dots, \v_n$ corresponding to eigenvalues $\lambda_1,\dots,\lambda_n.$ Let $$ P=(\v_1| \dots |\v_n) $$ be the $n\times n$ matrix whose columns are the eigenvectors. Then $$ P^{-1}AP=\left(\begin{array}{cccc} \lambda_1&0&\cdots&0\\0&\lambda_2&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n\end{array}\right), $$ the diagonal matrix with the eigenvalues down the main diagonal.

Let the matrix $A$ be $n\times n$ with $n$ linearly independent eigenvectors $\v_1,\dots, \v_n$ corresponding to eigenvalues $\lambda_1,\dots,\lambda_n.$ Let $$ P=(\v_1| \dots |\v_n) $$ be the $n\times n$ matrix whose columns are the eigenvectors. Then $$ P^{-1}AP=\left(\begin{array}{cccc} \lambda_1&0&\cdots&0\\0&\lambda_2&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n\end{array}\right), $$ the diagonal matrix with the eigenvalues down the main diagonal.

We know that an $n\times n$ matrix $A$ is diagonalisable if and only if $A$ has $n$ linearly independent eigenvectors.

Now say $\lambda_1,\dots,\lambda_m$ are distinct eigenvalues of $A$, with corresponding eigenvectors $\v_1,\dots,\v_m.$ Then we have also seen that $\v_1,\dots,\v_m$ are linearly independent.

Hence if $A$ is $n\times n$ with $n$ distinct eigenvalues, then $A$ is diagonalisable.

The question remains, if $A$ has fewer than $n$ distinct eigenvalues, how do we know if $A$ is diagonalisable?

Let $A=\left(\begin{array}{ccc}2&1&3\\0&1&0\\0&0&1\end{array}\right)$ and $B=\left(\begin{array}{ccc}2&1&3\\0&1&1\\0&0&1\end{array}\right)$.

Easy to see the characteristic equation of both $A$ and $B$ is $$(2-\lambda)(1-\lambda)^2=0,$$ so $\lambda=2,1,1.$

Let $A=\left(\begin{array}{ccc}2&1&3\\0&1&0\\0&0&1\end{array}\right)$ and $B=\left(\begin{array}{ccc}2&1&3\\0&1&1\\0&0&1\end{array}\right).$ Characteristic equation of both $A$ and $B$ is $(2-\lambda)(1-\lambda)^2=0,$ so $\lambda=2,1,1.$

Note that both matrices have two ($n\lt 3$) distinct eigenvalues.

For $A:$ We have that for $\lambda=2$

For $A:$ Now for $\lambda=1$

So $\;x_1=-x_2-3x_3$. Thus $\; \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ \end{array} \right) =\left( \begin{array}{c} -x_2-3x_3 \\ x_2 \\ x_3 \\ \end{array} \right) $ $ =x_2 \left( \begin{array}{c} -1 \\ 1 \\ 0 \\ \end{array} \right) + x_3 \left( \begin{array}{c} -3 \\ 0 \\ 1 \\ \end{array} \right). $

Therefore $A$ is diagonalisable, since $A$ has three linearly independent vectors.

Let $A=\left(\begin{array}{ccc}2&1&3\\0&1&0\\0&0&1\end{array}\right)$ and $B=\left(\begin{array}{ccc}2&1&3\\0&1&1\\0&0&1\end{array}\right).$ Characteristic equation of both $A$ and $B$ is $(2-\lambda)(1-\lambda)^2=0,$ so $\lambda=2,1,1.$

Now for $B:$ We have that for $\lambda=2$

Now for $B:$ On the other hand, for $\lambda=1$

Therefore $B$ is NOT diagonalisable, since $B$ has only two linearly independent vectors.

If we are only interested in finding out whether or not a matrix is diagonalisable, then we need to know the dimension of each eigenspace. There is one theorem (which we will not prove!) that states:

If $\lambda_i$ is an eigenvalue, then the dimension of the corresponding eigenspace cannot be greater than the number of times $(\lambda-\lambda_i)$ appears as a factor in the characteristic polynomial.

We often use the following terminology:

• The geometric multiplicity of the eigenvalue $\lambda_i$ is the dimension of the eigenspace corresponding to $\lambda_i$.
• The algebraic multiplicity of the eigenvalue $\lambda_i$ is the number of times $(\lambda-\lambda_i)$ appears as a factor in the characteristic polynomial.

The main result is the following:

16.5.1 Systems of differential equations

For a system of coupled differential equations which can be written in matrix form as $$ \dot{\mathbf x}=A{\mathbf x} $$ (where ${\mathbf x}=(x_1,\dots,x_n)^T,$ $\dot{\mathbf x}=(\dot x_1,\dots,\dot x_n)^T$), if $A$ can be diagonalised, say $P^{-1}AP=D$ with $D$ diagonal, then make the substitution ${\mathbf x}=P{\mathbf y}.$ This yields $$ \dot{\mathbf y}=D{\mathbf y} $$ which is easily solved.

16.5.1 Systems of differential equations

For example, consider the coupled system of ODEs: \[ \left\{ \begin{array}{rcrcr} \dot x_1 & = & x_1 & + & 2x_2\\ \dot x_2 & = & 2x_1 & + & x_2\\ \end{array} \right. \] Write $ \left( \begin{array}{r} \dot x_1 \\ \dot x_2 \\ \end{array} \right) = \left( \begin{array}{cc} 1 & 2\\ 2 & 1\\ \end{array} \right) \left( \begin{array}{r} x_1 \\ x_2 \\ \end{array} \right)\; $ or $\; \dot{\mathbf x} = A \mathbf x.$ $\quad (\star)$

Is $A$ diagonalizable? 🤔 The answer is "Yes". $\;\Ra \; P^{-1}A P = D.$

16.5.1 Systems of differential equations

Write $ \left( \begin{array}{r} \dot x_1 \\ \dot x_2 \\ \end{array} \right) = \left( \begin{array}{cc} 1 & 2\\ 2 & 1\\ \end{array} \right) \left( \begin{array}{r} x_1 \\ x_2 \\ \end{array} \right)\; $ or $\; \dot{\mathbf x} = A \mathbf x.$ $\quad (\star)$

Is $A$ diagonalizable? 🤔 The answer is "Yes". $\;\Ra \; P^{-1}A P = D.$

From $\,(\star)\,$ we have

16.5.1 Systems of differential equations

Introducing the expression $ \mathbf y(t) = \left( \begin{array}{r} y_1(t) \\ y_2(t) \\ \end{array} \right) = P^{-1}\mathbf x,\, $ implies

16.5.1 Systems of differential equations

Finally, solve for $\,y_1$ and $y_2,\,$ then substitute into $\,\mathbf x = P\mathbf y\,$ to get $\,x_1, x_2.$

If $A$ is diagonalisable, say $P^{-1}AP=D$ with $D$ diagonal, then $$ A^n=PD^nP^{-1}. $$ This gives an easy way to calculate $A^n$.

From $A= PDP^{-1}$ we have that

What can we say about the linear system (corresponding to a square matrix $A$)

$$ A{\mathbf x} = {\mathbf b}, $$

if we know how to diagonalise $A$?

Multiply $A{\mathbf x} = {\mathbf b}$ by $P^{-1}$ from left on both sides. Then we get

Set $\mathbf y = P^{-1}\mathbf x,$ then $D\mathbf y = P^{-1} \mathbf b.$ 👈 This is a decoupled system! Finally, solve for $\mathbf y $ and then $\mathbf x = P \mathbf y.$