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Does there exist a matrix which is both not invertible and not diagonalizable? If so, please provide an example.
Thanks,
David
Thanks,
David
HallsofIvy said:One that comes to mind immediately is
[tex]\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}[/tex]
A matrix is not invertible if and only if it has 0 as an eigenvalue. A matrix is diagonalizable if the exist a basis for the space consisting of eigenvectors. Those are not contradictory.
HallsofIvy said:Yes. So? A diagonal matrix is trivially "diagonalizable".
When a square matrix has a unique solution to the system of linear equations it represents, it is considered invertible. This means that it can be multiplied by another matrix, called its inverse, to give the identity matrix.
A matrix is invertible if its determinant is non-zero. This can be calculated by taking the product of the elements on the main diagonal and subtracting the product of the elements on the opposite diagonal.
No, only square matrices can be invertible. This is because non-square matrices do not have an equal number of rows and columns, making it impossible to find an inverse matrix that can multiply with it to give the identity matrix.
Diagonalizable matrices have a special property where they can be simplified into a diagonal matrix through a change of basis. This makes it easier to perform calculations and solve equations involving the matrix.
A matrix is diagonalizable if it has a full set of linearly independent eigenvectors. This means that the matrix can be decomposed into the product of a diagonal matrix and a matrix of eigenvectors.