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RE: I can't understand this matrix

$\begin{pmatrix}-1&1/3&0&0\\1&-1&2/3&0\\0&2/3&-1&1\\0&0&1/3&-1 \end{pmatrix}$

It has determinant 0, rank 3 out of a possible 4, and yet it has four linearly independent eigenvectors.

$\begin{pmatrix} 1\\3\\3\\1\end{pmatrix}$, $\begin{pmatrix} 1\\-1\\-1\\1\end{pmatrix}$, $\begin{pmatrix} -1\\-1\\1\\1\end{pmatrix}$,$\begin{pmatrix} -1\\3\\-3\\1\end{pmatrix}$

These are linearly independent, so they form a complete basis. I don't understand this. I thought that a matrix with determinant zero should not have a set of eigenvectors that make a complete basis. Where I am going wrong? I realize the first eigenvector has eigenvalue 0.

Thanks so much.

ps When I put these four eigenvectors into a matrix, it has rank 4, so I know they are linearly independent.

user73063
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  • It's not a problem that these are independent. It only means that the matrix is diagonalizable, hence that $\mathbb R^4$ is decomposed into a direct sum $E_0\oplus E_\lambda\oplus E_\mu\oplus E_\kappa$, where $\kappa,\mu,\lambda$ are the non-zero eigenvalues. – zarathustra May 07 '14 at 05:41
  • Thanks. But can a singular matrix (determinant 0) be diagonalizable? – user73063 May 07 '14 at 05:44
  • I mean, 1) a singular matrix cannot be inverted, 2) a matrix that is diagonalizable can be inverted. Where am I going wrong. – user73063 May 07 '14 at 05:44
  • Yes sure, in the appropriate basis your matrix is $\begin{pmatrix}\kappa & 0 & 0 & 0\0 & \mu & 0 & 0\0 & 0 & \lambda & 0\0 & 0 & 0 & 0\end{pmatrix}$ – zarathustra May 07 '14 at 05:45
  • So I guess my problem was thinking that being "diagonalizable" was the same as being "invertable". – user73063 May 07 '14 at 05:46
  • You certainly know that invertible means that the determinant is non-zero (in $\mathbb R$ or $\mathbb C$), and that the determinant is the product of the eigenvalues. Hence invertible means all eigenvalues are non-zero, whereas diagonalizable doesn't really care about what the eigenvalues are. – zarathustra May 07 '14 at 05:53
  • the zero matrix is not invertible, and is diagonlizable – Oria Gruber May 07 '14 at 05:59

2 Answers2

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As said in the comments,

It's not a problem that these are independent. It only means that the matrix is diagonalizable, hence that $\mathbb R^4$ is decomposed into a direct sum $E_0\oplus E_\lambda\oplus E_\mu\oplus E_\kappa$, where $\lambda,\mu,\kappa$ are the non-zero eigenvalues. Hence in the appropriate basis, your matrix is

$$\begin{pmatrix}\kappa & 0 & 0 & 0\\0 & \mu & 0 & 0\\0 & 0 & \lambda & 0\\0 & 0 & 0 & 0\end{pmatrix}$$

which is still not invertible, but is diagonal.

zarathustra
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An $n\times n$ matrix $A$ is diagonalizable if $A$ has $n$ linearly independent eigenvectors.

An $n\times n$ matrix $A$ is invertible if there exists an $n\times n$ matrix $B$ such that $AB=I_n$.

There are diagonalizable matrices that are not invertible. For example, the zero-matrix is diagonalizable but not invertible.

There are also invertible matrices that are not diagonalizable. For example, let $$ A= I_n+S_n $$ where $$ [S_n]_{ij}= \begin{cases} 0 & j\neq i+1 \\ 1 & j= i+1 \end{cases} $$ Then $A$ is invertible but not diagonalizable.

The two examples above demonstrate that diagonalizability is unrelated to invertibility.

Brian Fitzpatrick
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