# Finding characteristic matrix of a triangle matrix

I came to this stage when I was reading a Linear algebra text :

Suppose $M$ is a block triangular matrix, say $M= \begin{pmatrix} A_1 & B \\ 0 & A_2\end{pmatrix}$ where $A_1$ and $A_2$ are square matrices. Then the characteristic matrix of $M$, $$\begin{pmatrix} tI-A_1 & -B \\ 0 & tI-A_2\end{pmatrix}$$is also a block triangular matrix with diagonal blocks $tI-A_1$ and $tI-A_2$. Thus by Theorem 7.12, $$|tI-M|=\left|\begin{matrix} tI-A_1 & -B \\ 0 & tI-A_2\end{matrix}\right|=|tI-A_1||tI-A_2|.$$That is the characteristic polynomial of $M$ is the product of the characteristic polynomials of the diagonal blocks $A_1$ and $A_2$.

$tI - M$ gives a matrix with components :

• $tI - A_1$ (makes sense)
• $tI - A_2$ (even that I got)

But why is the top right component $-B$? Why not $tI-B$? What am I missing?

Note that $$M = \begin{pmatrix} A_1 & B \\ 0 & A_2 \end{pmatrix}$$ writing the identity with correspoding blocks: Suppose $A_1$ is a $n_1 \times n_1$-matrix and $A_2$ is a $n_2 \times n_2$-matrix. Then in $tI - M$ the identity is a $(n_1+n_2)\times (n_1 + n_2)$-matrix. We have $$I_{n_1+n_2} = \begin{pmatrix} I_{n_1} & 0 \\ 0 & I_{n_2} \end{pmatrix}$$ as the identity does only have off-diagonal entries and the off-diagonal blocks do not contain diagonal-elements. Hence $$tI - M = t\begin{pmatrix} I_{n_1} & 0 \\ 0 & I_{n_2} \end{pmatrix} - \begin{pmatrix} A_1 & B \\ 0 & A_2 \end{pmatrix} = \begin{pmatrix} tI - A_1 & -B \\ 0 & tI - A_2 \end{pmatrix}.$$

• The identity matrix! Yes! How did I not see that? Thanks a lot sir! – Macindows Jan 8 '17 at 17:22
• You're welcome. – martini Jan 8 '17 at 17:27