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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?

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

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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.

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

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Here is another quick solution:

1. we know that for any matrix $A \in M_{n}$, $$det(A) = \lambda_1 \times \lambda_2 \dots ... \lambda_n$$ where $\lambda_i$ are eigenvalues of $A$.

2. Also, we know a matrix is invertible iff its determinant is non-zero.
3. Finally, eigenvalues of an $n \times n$ triangular matrix is its diagonal entries.

Now, from above, we conclude that a triangular matrix $A$ is invertible iff all diagonal entries are non-zero. Otherwise, the product of them-which is equal to the detrminant- is zero and it will be singular.

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