# Why can the action of a diagonalizable matrix in a vector be written as linear combination of eigenvectors?

Suppose $$A \in \mathbb{C}^{d \times d}$$ is diagonalizable, with eigenvalues $$\lambda_1, ... , \lambda_d$$ and eigenvectors $$y_1,...,y_d$$. Define $$\Lambda = \operatorname{diag}(\lambda_1, ... , \lambda_d)$$ and $$P = (y_1,...,y_d)$$ unitary.

Since $$A$$ is diagonalizable, we know that $$A = P \Lambda P^{-1}$$.

So, suppose we have a vector $$x \in \mathbb{C}^d$$. Then $$Ax = P \Lambda P^{-1} x = \sum^d_{i=1} \lambda_1 c_i y_i$$, with $$c_i = y_i \cdot x$$

My question is: why is $$P \Lambda P^{-1} x = \sum^d_{i=1} \lambda_i c_i y_i$$, with $$c_i = y_i \cdot x$$?

It's not so clear to me why this is the case.

• @Azif00 $A$ is diagonalisable. We know that $Ay_i = y_i \lambda_i$, and so when we take all the eigenvectors and eigenvalues together, it becomes $A P = P \Lambda$, which will give the expression above. $\Lambda$ is the diagonal matrix. Is this correct? Commented Oct 19, 2019 at 4:15
• I guess your matrix is self-adjoint or normal. Otherwise this is false. Also, it should be $c_i = \overline{y_i}\cdot x$ in this case. Commented Oct 19, 2019 at 4:20
• @Azif00 I see. Also, I failed to mention we can take $P$ to be unitary. Commented Oct 19, 2019 at 4:25
• @14tim4 Yeah. I'm sorry. You are right, is $A=P\Lambda P^{-1}$. My apologies. Commented Oct 19, 2019 at 4:27

We can take the $$y_k$$ to be orthonormal (equivalently, $$P$$ is unitary). Then $$x = \sum_k (y_k^* x) y_k$$.
Hence $$Ax = \sum_k (y_k^* x) Ay_k = \sum_k \lambda_k(y_k^* x) y_k$$.