Is there any relationship between 'invertible' and 'diagonalizable'?

Question

I'am a linear algebra beginner and I am confusing with invertible and diagonalizable.

From my understanding, invertible means non-singular and any of eigenvalue must not be 0. Diagonalizable means there must be N linearly independent eigenvectors.(Eventhough eigenvalue has 0, it seems possible to have N linearly independent eigenvectors. Right?)

Above are just single fraction of their property, and I cannot imagine bigger picture than this.

Is there any intuitive relation or theorem between 'invertible' and 'diagonalizable'?

Answer 1

From my understanding, invertible means non-singular and any of eigenvalue must not be 0.

Exactly. In fact, a matrix is singular if and only if $0$ is its eigenvalue.

Diagonalizable means there must be N linearly independent eigenvectors.(Eventhough eigenvalue has 0, it seems possible to have N linearly independent eigenvectors. Right?)

Correct. Even if an eigenvalue is $0$, a matrix can have $N$ linearly independent eigenvectors. For example, the zero matrix has $N$ linearly independent eigenvectors, because every vector is an eigenvector for the zero matrix.

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Is there any intuitive relation or theorem between 'invertible' and 'diagonalizable'?

Not directly, in the sense that one would imply another. You can have matrices in all four classes, i.e.

• Invertible and diagonalizable. An example of this is the idenity matrix $\begin{bmatrix}1&0\\0&1\end{bmatrix}$.
• Invertible and not diagonalizable. The simples example would be $\begin{bmatrix}1&1\\0&1\end{bmatrix}$.
• Singular and diagonalizable. The zero matrix $\begin{bmatrix}0&0\\0&0\end{bmatrix}$. is the simplest example of this.
• Singular and non-diagonalizable. The simplest example is probably $\begin{bmatrix}0&1\\0&0\end{bmatrix}$.