Find a matrix $P$ that diagonalizes $\begin{bmatrix}0&0&-2\\1&2&1\\1&0&3\end{bmatrix}$



we found the following Eigenspaces:

$\lambda=2\\\lambda=1$ the eigenvalues and $P_1=\begin{bmatrix}-1\\0\\1\end{bmatrix},P_2=\begin{bmatrix}0\\1\\0\end{bmatrix},P_3=\begin{bmatrix}-2\\1\\1\end{bmatrix}$

There are three basis in total , so the matrix is diagonalizable and $P=\begin{bmatrix}-1&0&-2\\0&1&1\\1&0&1\end{bmatrix}$ diagonalizes $A$.


If you notice there are two equal eigenvalues with respect each of the matrix P columns. If you multiply PA.

Theorem : Let V be a vector space over $K$, and let $A:V\to V$ be an operator. Let $v_1,...,v_m$ be eigenvectors of $A$, with eigenvalues $\lambda_1,...,\lambda_m$ respectively. Assume that these eigenvalues are distinct,i.e.

$\lambda_i\neq\lambda_j$ if $i\neq j$

Then $v_1,...,v_m$ are linearly independent.


If you notice there are two equal eigenvalues with respect to each of the matrix columns P in the example. For the matrix to be invertible the columns must be linearly independent, which means by the theorem that need to have different eigenvalues. However 2 stands for the eigenvalue of two different columns of the matrix $P$. How can it be?

Thanks in advance!

  • $\begingroup$ A repeated eigen value can have the corresponding eigen space of dimension $>1$. Thus you can have more than one linearly independent (eigen) vectors corresponding to that eigen value. Just think of the identity matrix, all its eigen values are $1$, take a guess what can be an eigen vector....? $\endgroup$
    – Anurag A
    Sep 10, 2017 at 20:23

1 Answer 1


The theorem says that eigenspaces corresponding to distinct eigenvalues are linearly independent. It says nothing about the converse.

In your example, the eigenvalue $2$ has multiplicity two, and that's why you can find two linearly independent eigenvectors. 'That's very easy to see in the diagonalized version of your matrix: $$ \begin{bmatrix} 2&0&0\\0&2&0\\0&0&1\end{bmatrix}. $$


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