# Why can't a triangular matrix with only zeros in its diagonal be invertible?

Why can't a triangular matrix with only zeros in its diagonal be invertible?

I know that it is not invertible but I don't know well the reasons, perhaps. Actually, I read that can't have any zero in its diagonal, in order to be invertible.

In general, I know that if a matrix has an inverse, the product of that matrix with its inverse is equal to the identity matrix, namely, with a matrix with 1s as its diagonal.

• Can you state precisely, which matrix are you talking about. The first sentence is misleading – Ulysses Nov 4 '15 at 8:21
• hollow matrix, it is called, if I am not wrong. A matrix which has 0s in its diagonal. – Always learning Nov 4 '15 at 8:22
• @Ulysses, sorry, you right. I mean a triangular matrix with 0s in its diagonal. – Always learning Nov 4 '15 at 8:31
• Laplace expansion can yield that such matrix determinant is zero, thus it's uninvertible. en.wikipedia.org/wiki/Laplace_expansion – mbaitoff Nov 4 '15 at 8:40

A upper triangular matrix with 0's on its diagonal has its first column filled with 0's. Therefore its determinant is 0, which means it is not invertible. For a lower triangular matrix, the same holds with the last column.

In case you're not familiar with determinants:When A is upper triangular, let's suppose $A^{-1}$ exists. Then $A^{-1}A=I$. Let $B=\begin{pmatrix} 1 \\ 0\\0\\...\\0 \end{pmatrix}$. Then $AB=0$. However, $(A^{-1}A)B=I B=B$ , but $$A^{-1}(AB)=A^{-1}0=0\ne B$$ We have a contradiction, therefore A is not invertible.

• Can we find another way to state it, without referring to the determinant? Just because I am not sure if the determinant was already treated in the course. PS. I mean, I am almost sure that it was not introduced yet the determinant. – Always learning Nov 4 '15 at 8:40
• Edited the answer – GBQT Nov 4 '15 at 8:43

(The original question did not ask for a triangular matrix)

There are (many) invertible matrices with a zero diagonal, for example consider the matrix $$A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$ then $$A^2 = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$$ so $A$ is invertible with $A^{-1} = A$.

Addendum: For the reformultated question, note that a triangular matrix with zero diagonal, has a zero column. Hence its kernel is not trivial, therefore it is not invertible.

• Sorry, @Ulysses was right. I did not specify that I refer to a triangular matrix. – Always learning Nov 4 '15 at 8:30
• @Alwayslearning Corrected. – martini Nov 4 '15 at 8:37

This is not true, as $\left(\begin{array}{c c}0 & 1\\ 1 & 0\end{array}\right)$ is invertible.

• Sorry, @Ulysses was right. I did not specify that I refer to a triangular matrix. – Always learning Nov 4 '15 at 8:30

Here is the best version I could think of completely avoiding advanced concepts like kernels, determinants and the like.

Any matrix that has a row or column which is a zero vector cannot be inverted. If we have $$A= \begin{pmatrix} -&a_1&-\\ \vdots&\vdots&\vdots\\ -&a_n&- \end{pmatrix} \text{ and } B= \begin{pmatrix} |&\cdots&|\\ b_1&\cdots&b_n\\ |&\cdots&| \end{pmatrix}$$ with $a_i=0$ for some $i$, then $a_i\cdot b_i=0$ making entry $(A\cdot B)_{i,i}=0$, so their product cannot be the identity matrix. Knowing that inverses are always two-sided, similar considerations work for the column vector case.

Now a triangular matrix with zeros on its main diagonal has both a zero row and a zero column, and the statement follows.