So I was doing a matrix problem that required the use of a particular theorem, which states that,

" An $n\times n$ matrix $A$ is diagonalisable if and only if $\exists$ an invertible $P$ and diagonal matrix $D$ such that $P^{-1}AP=D$, where

$(1)$ The columns of $P$ are $n$ linearly independent eigenvectors of $A$;

$(2)$ The diagonal entries of $D$ are the eigenvalues of $A$ corresponding to said eigenvectors of $A$, which are the columns of $P$ in the respective order. "

What particularly interests me with this theorem are its last few words - "in the respective order".

I understand that, if we consider the case of having unique diagonal entries in $D$ (i.e. eigenvalues of $A$ with algebraic multiplicities of $n=1$), then upon finding the eigenvectors for the $i^{th}$ diagonal entry $\lambda_i$ of $D$ (where $i$ is the entry's column or row number) by reducing the augmented matrix $[A-\lambda_i\mathbb{I}|0]$, then we will end up with a general eigenvector with $n=1$ free variables $\in \mathbb{R}$, which we can then factor out to give us a basis for the corresponding eigenspace $E_\lambda$. Thus, the $i^{th}$ eigenvalue will have an eigenvector equal to the $i^{th}$ column of $A$, which will be a scalar multiple of the vector in the basis. And so we can construct $P$ using the theorem by fixing the $i^{th}$ columns of $P$ respectively as eigenvectors corresponding to $\lambda_i$.

Similarly, in the case of $D$ having diagonal elements which are not all unique (i.e some eigenvalues have algebraic multiplicities $n>1$), then upon finding the eigenvectors for a $\lambda$ with algebraic multiplicity $m>1$ by using the same method as above, we will end up with a general eigenvector with $m$ free variables which we can then split up into a basis of $m$ vectors for the eigenspace $E_\lambda$ (for the sake of the question we assume that our matrix $A$ is diagonalisable and thus each eigenvector is linearly independent). This means, however, that when constructing $P$, unlike in the previous case, we cannot simply fix the $i^{th}$ columns of $P$ (where $i$ is the $m$ different row or column numbers in which $\lambda$ appears) as the corresponding $m$ eigenvectors for $\lambda$ as there is now more than one way or combination to do this in; these $m$ eigenvectors were not derived in an immediately noticeable order, unlike their corresponding eigenvalues which are, and thus pairing eigenvector to column becomes trickier.

Playing about, I thought I found the right approach however. Consider the $4\times4$ diagonalisable matrix, $A$, below.

$A=\left[\begin{array}{cccc} 9 & 0 & 0 & 18 \\ 0 & 9 & 0 & 0 \\ 0 & 0 & -9 & 0 \\ 0 & 0 & 0 & -9 \end{array}\right]$.

Since $A$ is triangular, its eigenvalues are the entries along its main diagonal. Therefore, the eigenvalues of $A$ are $\lambda=9$ and $\lambda=-9$.

Substituting $\lambda=9$ into the augmented matrix $[A-\lambda\mathbb{I}|0]$ to solve for corresponding eigenvectors, we get

$\left[\begin{array}{rrrr|r} 0 & 0 & 0 & 18 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -18 & 0 & 0 \\ 0 & 0 & 0 & -18 & 0 \end{array}\right]$.

So we have $x_1=x_1, x_2=x_2, x_3=0$ and $x_4=0$. Then our eigenvector

$\underline{x}=\left[\begin{array}{rrrr} x_1 \\ x_2 \\ 0 \\ 0 \end{array}\right]=x_1\left[\begin{array}{rrrr} 1 \\ 0 \\ 0 \\ 0 \end{array}\right]+x_2\left[\begin{array}{rrrr} 0 \\ 1 \\ 0 \\ 0 \end{array}\right]$ for any $x_1, x_2\in\mathbb{R}$.

Similarly for $\lambda=-9$ we get:

$\left[\begin{array}{rrrr|r} 18 & 0 & 0 & 18 & 0 \\ 0 & 18 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]$.

So we have $x_1=-x_4, x_2=0, x_3=x_3$ and $x_4=x_4$. Then our eigenvector

$\underline{x}=\left[\begin{array}{rrrr} -x_4 \\ 0 \\ x_3 \\ x_4 \end{array}\right]=x_3\left[\begin{array}{rrrr} 0 \\ 0 \\ 1 \\ 0 \end{array}\right]+x_4\left[\begin{array}{rrrr} -1 \\ 0 \\ 0 \\ 1 \end{array}\right]$ for any $x_3, x_4\in\mathbb{R}$.

I then noticed that each eigenvector had, together, perfectly factored out the four different variables $x_1,x_2,x_3,x_4$.

I imagined that this was the rule; that each eigenvector's factored out variable number would determine the eigenvector's column number in $P$, i.e. the eigenvector with $x_n$ factored out would be positioned at the $n^{th}$ column. Thus, I constructed

$P=\left[\begin{array}{cccc} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$

and sure enough, when using the eigenvalues of $A$ to construct $D$, I reached the equality $P^{-1}AP=D$, as required by the theorem.

So my question is (finally):

In the case of diagonalisable matrices having non-unique eigenvalues, is the 'rule' mentioned above the right way to proceed in constructing $P$? It worked in this example but is it concrete for all cases?

Thanks in advance!


The only thing you need is for the $i$-th column of $P$ (call it $p_i$) to be an eigenvector with the eigenvalue of $\lambda_i$, where $\lambda_i$ is the $i$-th value of the diagonal.

There is no demand that $\lambda_i$ is unique. As long as $Ap_i=\lambda_i p_i$ for all $i$, and the columns $p_1,\dots, p_n$ are independent, you are good to go.

Naturally, if an eigenvalue has geometric multiplicity $\geq 2$, then $\lambda_i=\lambda_{i+1}$, which means that $p_i, p_{i+1}$ must be two independent elements of $\ker(A-\lambda_i I)$, but their order is irrelevant.


The characteristic polynomial of a square matrix sometimes has multiple roots. There is a divisor called the minimal polynomial, which is the $q(x)$ of lowest degree such that $q(A) = 0.$

The matrix $A$ is diagonalizable if and only if the minimal polynomial is squarefree. As soon as $q(x)$ has a repeated root, the Jordan form has some nontrivial blocks for that eigenvalue. If the eigenvalues are really messy, care must be used in finding $q(x),$ although its coefficients will be nice. Finally, multiple roots may be detected if $\gcd( q(x), q'(x)) \neq 1$


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