Diagonal, Jordan Normal and And Inverses

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SUMMARY

A diagonal matrix is not always invertible; it requires all diagonal entries to be nonzero for its inverse to exist. An invertible matrix is not necessarily diagonalizable, as demonstrated by the matrix A = [[1, 0], [1, 1]], which is invertible but lacks sufficient linearly independent eigenvectors. The invertibility of a matrix in Jordan normal form is not guaranteed, as shown by the example matrix [[1, 0, 0], [0, 1, 0], [0, 0, 0]], which is not invertible due to a zero row.

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1) Is a diagonal matrix always invertable?
2) Is an Invertable matrix always Diagonalizable?
3) Is a matrix in jordan normal form always invertable

The answers are prob straight foward but I am confused.
 
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Jordan Normal, Diagonal and Inverses

Homework Statement



1) Is a diagonal matrix always invertable?
2) Is an Invertable matrix always Diagonalizable?
3) Is a matrix in jordan normal form always invertable

The answers are prob straight foward but I am confused.

Homework Equations





The Attempt at a Solution

 
I've had a thought about this, I guess what I am looking for is some sort of relation between the characteristic polinomial and the determinant of a the matrix.
 
1) No. Let d_{i} be the diagonal entries of a diagonal matrix The inverse (if it exists) consists of the values \frac {1}{d_{i}} along the diagonal. Therefore, all d_{i}'s must be nonzero for the inverse to exist.

2) No. If an nxn matrix is diagonalizable, it must have n linearly independent eigenvectors. A = \left(\begin{array}{cc}1&0\\1&1\end{array}\right) is invertibile but not diagonalizable.

3) I don't know. :rolleyes:
 


Is
\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 0\end{bmatrix}
invertible?
 
Thanks greatly.

That clears it all up.
 

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