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user371838
user371838

Let A$A$ have an eigenvalue of 0$0$. By definition, there is a non-trivial vector v$v$ such that Av = 0v = 0$Av = 0v = 0$. Let B$B$ be any matrix. Then BA$BA$ is not the identity matrix, because (BA)v = B (Av) = B0 = 0.$(BA)v = B (Av) = B0 = 0.$ This is true for any matrix B$B$, so A$A$ is not invertible.

Let A have an eigenvalue of 0. By definition, there is a non-trivial vector v such that Av = 0v = 0. Let B be any matrix. Then BA is not the identity matrix, because (BA)v = B (Av) = B0 = 0. This is true for any matrix B, so A is not invertible.

Let $A$ have an eigenvalue of $0$. By definition, there is a non-trivial vector $v$ such that $Av = 0v = 0$. Let $B$ be any matrix. Then $BA$ is not the identity matrix, because $(BA)v = B (Av) = B0 = 0.$ This is true for any matrix $B$, so $A$ is not invertible.

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gnasher729
gnasher729
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Let A have an eigenvalue of 0. By definition, there is a non-trivial vector v such that Av = 0v = 0. Let B be any matrix. Then BA is not the identity matrix, because (BA)v = B (Av) = B0 = 0. This is true for any matrix B, so A is not invertible.