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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.

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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 $A$?

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  • $\begingroup$ 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? $\endgroup$ – Jaigus Mar 13 at 18:43
  • $\begingroup$ @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 $A_{2,2}$. $\endgroup$ – Morgan Rodgers Mar 13 at 19:30
  • $\begingroup$ 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? $\endgroup$ – Jaigus Mar 13 at 19:59
  • $\begingroup$ Yes absolutely, a full column of zeros will give zero determinant with a cofactor expansion also. $\endgroup$ – Morgan Rodgers Mar 13 at 20:24
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    $\begingroup$ @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 $\begin{bmatrix} 0 & A_{2,2}\end{bmatrix}$? $\endgroup$ – Morgan Rodgers Mar 13 at 23:03
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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.

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