I saw one operation (stated below) in a proof and I don't think I completely understand its innate operation logic although I can guess the answer.
The proof idea is:
Suppose $\exists$ a non-singular matrix $S_{n \times n}$ s.t. $$A = S([\lambda]\oplus B) S^{-1} $$
Let x $\in$ $\mathbb{R}^n$ be the 1st column of S, so partition $$S = [x \quad S_1]$$
Then $$ AS = A[x \quad S_1] = S([\lambda] \oplus B) = [x \quad S_1]([\lambda] \oplus B) = [\lambda x \quad S_1 B] $$
What I don't understand is what operation rule does the last equation follow.
From the definition of Kronecker sum, $$ [\lambda] \oplus B = \lambda \otimes I_{n-1} + I_{1} \otimes B $$, which is an $(n-1)\times (n-1)$ matrix, then it looks dimensionally incompatible between the partitioned matrix $S = [x \quad S_1]$ and this $(n-1)\times (n-1)$ Kronecker sum matrix $\lambda \otimes I_{n-1} + I_{1} \otimes B$.
Can anyone explain this?
Thanks in advance!
So the $\oplus$ in $\lambda \oplus B$ is NOT a Kronecker sum, but a direct sum. This means $\lambda \oplus B$ is essentially a block diagonal matrix with the 1st diagonal block having the size $1 \times 1$ and with element $\lambda$ and the remaining diagonal block is B which is of the size $(n-1) \times (n-1)$. And all the off-diagonal blocks are zero blocks.