# How can I show that this matrix has no inverse?

Let $$A = [a_{ij}]$$ be an $$n \times n$$ matrix with entries in $$\mathbb{R}$$. Suppose there exists an $$m$$ with $$a_{ij} = 0$$ for $$i \ge m$$ and $$j \le m$$, and $$a_{i,i} \ne 0$$ for $$1 \le i \lt m$$. Show that $$A$$ has no inverse.

From my understanding no diagonal value is zero, but all the non-diagonal values more than some arbitrary $$m$$ is zero. And, I know I must show that the determinant is zero, but I'm not sure how to do this.

Note that you do have a diagonal entry equal to 0: $$a_{m,m} = 0$$. And possibly $$a_{k,k} = 0$$ for values of $$k \geq m$$.

The hypothesis is saying you have a block of zeroes in the bottom left of the matrix, which includes a diagonal entry. This should allow you to take a block decomposition of your matrix, $$A = \begin{bmatrix} A_{1,1} & A_{1,2} \\ 0 & A_{2,2} \end{bmatrix}$$ where $$A_{1,1}$$ is $$(m-1) \times (m-1)$$ and has no zeroes on the diagonal. Then $$A_{2,2}$$ is $$(n-(m+1))\times (n-(m+1))$$, the top left entry of $$A_{2,2}$$ is $$a_{m,m}$$ (which is 0). In fact, the first column of $$A_{2,2}$$ consists of the entries $$a_{i,j}$$ where $$i$$ runs through $$m, m+1, \dots, n$$ and $$j = m$$ (your hypothesis gives you some info on these entries).

Now: Can you show that $$A_{2,2}$$ is not invertible? What does this tell you about the invertibility of $$A$$?

• If $A_{2,2}$'s top-left element is zero, then it's determinant should also be zero after taking the co-factor expansion. However, can't we take a non-singular matrix, and generate a zero in that very place by using row operations? – Jaigus Mar 13 at 18:43
• @Jaigus The comment you made is not true: simply having a diagonal entry equal to $0$ does not make the determinant $0$ (consider $\begin{bmatrix} 0 & 1\\1&0\end{bmatrix}$). But consider what is happening with the rest of the entries in the first column of $A_{2,2}$. – Morgan Rodgers Mar 13 at 19:30
• So because the first column is all zeros, that means that the first column as a vector is linearly dependent, so $A_{2,2}$ is not full rank, and therefore singular? – Jaigus Mar 13 at 19:59
• Yes absolutely, a full column of zeros will give zero determinant with a cofactor expansion also. – Morgan Rodgers Mar 13 at 20:24
• @Jaigus It's really the singularity of $A_{2,2}$ along with the $0$ block that will tell you the whole thing is singular. Have you seen formulas for the determinant of $\begin{bmatrix} A&B\\0&D \end{bmatrix}$? If not, consider the rank of $A$. How many pivot positions can $A$ have in the top portion of the matrix $\begin{bmatrix} A_{1,1} & A_{1,2}\end{bmatrix}$? How many in the bottom portion $\begin{bmatrix} 0 & A_{2,2}\end{bmatrix}$? – Morgan Rodgers Mar 13 at 23:03

The determinant $$\det(A)$$ is a sum of products $$\pm\, a_{1\,\sigma(1)}\>a_{2\,\sigma(2)}\>\cdots\> a_{n\,\sigma(n)}\ ,\tag{1}$$ where $$\sigma$$ runs through the $$n!$$ permutations of $$[n]$$. For such a permutation $$\sigma$$ we cannot have $$m+1\leq \sigma(i)\leq n\quad(m\leq i\leq n)$$ (counting elements). It follows that in each product $$(1)$$ there is at least one factor $$a_{i\,\sigma(i)}=0$$. Therefore $$\det(A)=0$$; hence $$A$$ is singular.