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Could anyone help me with this proof without using determinant? I tried two ways.

Let $A$ be a matrix. If $A$ has the property that each row sums to zero, then there does not exist any matrix $X$ such that $AX=I$, where $I$ denotes the identity matrix.

I then get stuck. The other way was to prove by contradiction, and I failed too.

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Hint: You can sum the elements of a row by multiplying this row with a vector of $1$'s. Can you find now a matrix $X$ (with appropriate columns) such that $AX=Ο$?

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    $\begingroup$ OMG how clever are you! Thank you very much! $\endgroup$ – Andy Z Nov 11 '15 at 14:12
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    $\begingroup$ Nice. Took me a sec :) $\endgroup$ – nbubis Nov 11 '15 at 15:18
  • $\begingroup$ $X$ need not be a matrix. $X$ could be a non-zero vector. (It's the obvious one.) $\endgroup$ – Eric Towers Nov 12 '15 at 0:01
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    $\begingroup$ Once there is$AX=0$, we can find the equation with contradiction: $A^{-1}AX=A^{-1}0$, where $LHS=X\neq LHS=0$. Nice method.;) $\endgroup$ – Chris Tang Dec 1 '19 at 5:52
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    $\begingroup$ Based on the property of invertible matrices, $Ax=0$ should only have the trivial solution; however, we find $Ax=0, x=(1,1,...,1)$ besides $x=\vec{0}$, which also proves $A$ is not invertible. $\endgroup$ – Chris Tang Dec 2 '19 at 8:49
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If the sum of the rows is zero, then the matrix has the eigenvalue $0$. As a result its $\ker$ is of dimension $\ge1$, i.e. there is a nonzero solution to $AX=0$, hence it's noninvertible

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  • $\begingroup$ Well as I mentioned, it's better to avoid anything related to determinant. It is an exercise question after matrix multiplication, so I think we don't really need deep concepts to prove this. $\endgroup$ – Andy Z Nov 11 '15 at 14:11
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    $\begingroup$ @H.Zhu Eigenvalues don't need to be related to determinants. There are several ways to define them without using determinants and characteristic polynomials $\endgroup$ – Hippalectryon Nov 11 '15 at 14:14
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Assuming $A$ is an $n \times n$ matrix, let $v_0 = (0,0,\dots,0)$ and $v_1 = (1,1,\dots,1)$ be $n$-element column vectors. Since each row of $A$ sums to zero, it follows that $$A v_1 = (0,0,\dots,0) = A v_0,$$ showing that $A$ cannot have a (left) inverse.

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If each row of $A$ sums to zero, then each row of the column vector that is the sum of the column vectors constituting $A$ is zero. So the columns of $A$ are not linearly independent, and therefore the matrix is singular (i.e. it has no inverse).

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Hint: For a matrix $A$ having such a property has vector $(1,1,...,1)$ in its kernel thereby giving $\dim(\ker A)>0$.

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Theorem. If $A$ is an $n$-by-$n$ matrix, then $A$ is not invertible if and only if zero is an eigenvalue of $A$.

If $e$ denotes the all-ones vector of appropriate size, then, by hypothesis, $Ae = 0 = 0e$, i.e., zero is an eigenvalue of $A$.

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I'm not too sure but my reasoning is that if the sum of a row is 0, then the rows of the matrix A are linearly dependent because they are a linear combination. If the rows of A are linearly dependent then the columns of A transpose is linearly dependent. Therefore matrix A is not invertible by the Invertible Matrix Theorem and the determinant is equal to 0.

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