### Counterexample

We give a counterexample. Consider the \$2 imes 2\$ zero matrix.The zero matrix is a diagonal matrix, and thus it is diagonalizable.However, the zero matrix is not invertible as its determinant is zero.

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### More Theoretical Explanation

Let us give a more theoretical explanation.If an \$n imes n\$ matrix \$A\$ is diagonalizable, then there exists an invertible matrix \$P\$ such thatwhere \$lambda_1, dots, lambda_n\$ are eigenvalues of \$A\$.Then we consider the determinants of the matrices of both sides.The determinant of the left hand side iseginalign*det(P^-1AP)=det(P)^-1det(A)det(P)=det(A).endalign*On the other hand, the determinant of the right hand side is the productsince the right matrix is diagonal.Hence we obtain(Note that it is always true that the determinant of a matrix is the product of its eigenvalues regardless diagonalizability. See the post “Determinant/trace and eigenvalues of a matrix“.)

Hence if one of the eigenvalues of \$A\$ is zero, then the determinant of \$A\$ is zero, and hence \$A\$ is not invertible.

The true statement is:

### Is Every Invertible Matrix Diagonalizable?

Note that it is not true that every invertible matrix is diagonalizable.

For example, consider the matrixThe determinant of \$A\$ is \$1\$, hence \$A\$ is invertible.The characteristic polynomial of \$A\$ iseginalign*p(t)=det(A-tI)=eginvmatrix 1-t & 1\ 0& 1-tendvmatrix=(1-t)^2.endalign*Thus, the eigenvalue of \$A\$ is \$1\$ with algebraic multiplicity \$2\$.We haveand thus eigenvectors corresponding to the eigenvalue \$1\$ arefor any nonzero scalar \$a\$.Thus, the geometric multiplicity of the eigenvalue \$1\$ is \$1\$.Since the geometric multiplicity is strictly less than the algebraic multiplicity, the matrix \$A\$ is defective and not diagonalizable.

### Is There a Matrix that is Not Diagonalizable and Not Invertible?

Finally, note that there is a matrix which is not diagonalizable and not invertible.For example, the matrix \$eginbmatrix 0 & 1\ 0& 0endbmatrix\$ is such a matrix.

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## Summary

There are all possibilities.

Diagonalizable, but not invertible.Example: <eginbmatrix 0 & 0\ 0& 0endbmatrix.>Invertible, but not diagonalizable.Example: <eginbmatrix 1 & 1\ 0& 1endbmatrix>Not diagonalizable and Not invertible.Example: <eginbmatrix 0 & 1\ 0& 0endbmatrix.>Diagonalizable and invertibleExample: <eginbmatrix 1 & 0\ 0& 1endbmatrix.>Click here if solved 72