# Invertible matrix and different eigenvalues for each columns

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

Solution:

$$(\lambda-1)(\lambda-2)^2=0$$

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 $$A$$.

$$P^{-1}AP=\begin{bmatrix}-1&0&-2\\0&1&1\\1&0&1\end{bmatrix}\begin{bmatrix}0&0&-2\\1&2&1\\1&0&3\end{bmatrix}\begin{bmatrix}-1&0&-2\\0&1&1\\1&0&1\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&1\end{bmatrix}$$

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.

Question:

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?

• 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....? Sep 10, 2017 at 20:23

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: $$\begin{bmatrix} 2&0&0\\0&2&0\\0&0&1\end{bmatrix}.$$