Prove that a matrix in NOT invertible if an eignvalue is 0

**ishan** · December 7th 2008, 02:33 AM

^^

Thank you very much!

**Moo** · December 7th 2008, 02:44 AM

Originally Posted by ishan

^^

Thank you very much!

In the basis that diagonalizes the matrix, the matrix can be written :

$\begin{pmatrix} \lambda_1 & 0 & 0 &\ldots & 0 \\ 0 & \lambda_2 & 0 & \ldots & 0 \\ 0 & 0 & \lambda_3 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & 0 \\ 0 & 0 & 0 & \ldots & \lambda_n \end{pmatrix}$

Where $\lambda_1, \dots, \lambda_n$ are the eigenvalues of the matrix.
The determinant in this base is obviously the product of the $\lambda_i$
So if one of the eigenvalues is 0, the determinant is 0.
And since the determinant doesn't change with the basis, we have proved that det A = 0, and hence it is not invertible.

Edit : waaaah 39

th post

**whipflip15** · December 7th 2008, 03:01 AM

The determinant of a matrix is equal to the product of the eigenvalues. Hence the determinant is 0 and the matrix is not invertible.

**ThePerfectHacker** · December 7th 2008, 08:18 AM

If $A$ has $0$ as an eigenvalue then it means $A\bold{x} = 0\bold{x}$ for $\bold{x} \not = \bold{0}$ .
Therefore, $A\bold{x} = \bold{0}$ for $\bold{x}\not = \bold{0}$ .
This means $A$ is not invertible.

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