^^

Thank you very much!

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- Dec 7th 2008, 02:33 AMishanProve that a matrix in NOT invertible if an eignvalue is 0
^^

Thank you very much! - Dec 7th 2008, 02:44 AMMoo
In the basis that diagonalizes the matrix, the matrix can be written :

$\displaystyle \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 $\displaystyle \lambda_1, \dots, \lambda_n$ are the eigenvalues of the matrix.

The determinant in this base is obviously the product of the $\displaystyle \lambda_i$

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 :p - Dec 7th 2008, 03:01 AMwhipflip15
The determinant of a matrix is equal to the product of the eigenvalues. Hence the determinant is 0 and the matrix is not invertible.

- Dec 7th 2008, 08:18 AMThePerfectHacker
If $\displaystyle A$ has $\displaystyle 0$ as an eigenvalue then it means $\displaystyle A\bold{x} = 0\bold{x}$ for $\displaystyle \bold{x} \not = \bold{0}$.

Therefore, $\displaystyle A\bold{x} = \bold{0}$ for $\displaystyle \bold{x}\not = \bold{0}$.

This means $\displaystyle A$ is not invertible.