# Rank of upper triangular block with Identity matrix

Quick question I was hoping I could get some insight on. If I have a block matrix of the following form

$$A = \begin{pmatrix} I & A_{12} \\ 0 & A_{22} \end{pmatrix}$$

Where $A_{12}, A_{22}$ are arbitrary block matrices that need not be square (in fact in my case they are not square). Is it correct to say that

$$\text{rank} A = \text{rank}I + \text{rank}A_{22}$$

Because the identity matrix ensures that $A$ has the rank of $I$ and then the rank of $A$ should be determined by the rank of $A_{22}$, no?

• Yes, you are right. Nov 27, 2017 at 16:07
• Is there a more formal way that I can verify this or do you think it should be clear from here? Nov 27, 2017 at 16:15

Yes, your statement is correct. For a slightly more formal justification, note that $$\pmatrix{I&A_{12}\\0&A_{22}} \pmatrix{I & -A_{12}\\0&I} = \pmatrix{I&0\\0&A_{22}}$$ has total rank $\operatorname{rank}(I) + \operatorname{rank}(A_{22})$.

This isn't as elegant as the answer of Omnomnomnom but an other way to prove the statement.

Let be $I\in\mathbb C^{k\times k}$, $B\in\mathbb C^{k\times m}$ and $C\in\mathbb C^{n\times m}$ consequently $0\in\mathbb C^{n\times k}$. We consider $$A=\begin{pmatrix}I & B\\0 & C\end{pmatrix}$$ The rank is given by the maximal number of linearly independent rows (or columns) of $A$.

We denote $b_p$ as the $p$-th row of $B$ and $c_p$ as the $p$-th row of $C$.

Let be $\gamma=rank(C)$. Then there exists $\gamma$ rows $c_{p_1},\ldots,c_{p_\gamma}$ which are linear independent. Now consider $$\alpha_1\begin{pmatrix}1\\0\\\vdots\\0\\b_1^T\end{pmatrix} + \ldots+ \alpha_k\begin{pmatrix}0\\\vdots\\0\\1\\b_k^T\end{pmatrix}+\alpha_{k+1} \begin{pmatrix}0\\0\\\vdots\\0\\c_{p_1}^T\end{pmatrix} +\ldots+ \alpha_{k+\gamma}\begin{pmatrix}0\\\vdots\\0\\0\\c_{p_\gamma}^T\end{pmatrix}=0$$ You see directly that $\alpha_1=\ldots=\alpha_k=0$ and since $c_{p_1},\ldots,c_{p_\gamma}$ linear independent, you get further $\alpha_{k+1}=\ldots=\alpha_{k+\gamma}=0$. Hence $rank(A)\geq k+\gamma=rank(I)+rank(C)$.

Suppose $rank(A)>k+\gamma$, then you get at least $k+\gamma+1$ linear independent rows of $A$. But at least $\gamma+1$ rows are from the last $n$ rows. Since the first $k$ components are $0$, you can drop them and get $\gamma+1$ linear independent rows of $C$. This is a contradiction since $C$ contains at most $\gamma$ linear independent rows.

Together you get $rank(A)=k+\gamma=rank(I)+rank(C)$.

• This is also a very nice way to look at. Thanks for the alternate perspective! Nov 27, 2017 at 16:51