# Prove that an upper triangular matrix is invertible if and only if every diagonal entry is non-zero.

Prove that an upper triangular matrix is invertible if and only if every diagonal entry is non-zero.
I have proved that if every diagonal entry is non-zero, then the matrix is invertible by showing we can row reduce the matrix to an identity matrix. But how do I prove the only if part?

• Do you know the permutation based definition of a determinant? – Apurv May 1 '15 at 11:41
• No.. I don't think I can use determinant in this proof – In78 May 1 '15 at 16:07

If $A$ is an $n\times n$ triangular matrix, consider the system of equations $$A\mathbf x=\mathbf 0$$

If last $0$ in the main diagonal is at position $j$, you can solve for $x_n$, $x_{n-1}$,...,$x_{j+1}$. But what happens with $x_j$? Must it be $0$?

But if $A$ had an inverse we would have $$A^{-1}A\mathbf x=\mathbf x=\mathbf 0$$

Can you complete the reasoning?

If not every diagonal entry is zero, then we can show that the matrix does not have a full rank anymore (can you do it?). Therefore it is not invertible.

• Perhaps you prefer edit your deleted answer and then undelete it :) – ajotatxe May 1 '15 at 11:54
• @ajotatxe I would. But deleted answers do not appear on the IOS application. They should fix this. – user230734 May 1 '15 at 11:55
• Ok, sadly I can't do it for you because I can not undelete your post. – ajotatxe May 1 '15 at 12:00

Proof. ($\Rightarrow$)

Let $A=[a_{i,j}]_{n\times n}$ be an invertible upper triangular matrix. Suppose that $A$ has a diagonal entry that is zero, i.e., $a_{k,k}=0$, where $k\in\mathbb{N}$, $1\leq k<n$ (note that if $k=n$, then A would have a zero row, thus making A singular, which should not be the case). Then, the homogenous equation $Ax=0$, i.e.,

$$\left[ \begin{array}{ccc} a_{1,1} &&\cdots &&a_{1,n}\\ 0 &\ddots\\ \vdots & &a_{k,k} &&\vdots\\ &&&\ddots\\ 0&&\cdots&0&a_{n,n} \end{array} \right]\cdot \left[ \begin{array}{c} x_1\\ \vdots\\ x_k\\ \vdots\\ x_n \end{array}\right]=0,$$

where the $k^{th}$ row is the last row whose entry on the main diagonal is zero, will have a nontrivial solution since

$$\underbrace{\overbrace{a_{k-1,k-1}x_{k-1}}^{\text{if k\neq1}}+a_{k-1,k}x_k}_{(i)}+\underbrace{a_{k-1,k+1}x_{k+1}+\cdots+a_{k-1,n}x_n}_{equals\text{ }to\text{ }0}=0,$$

wherein (i) implies that we have, indeed, a nontrivial solution. Thus, a contradiction exists since if $A$ is invertible, then the homogenous equation $Ax=0$ must only have the trivial solution.

Therefore, all of the diagonal entries of $A$ must be nonzero.