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Question:

How would I go about proving that some square matrix $A$ is invertible iff $\det{A} ≠ 0$?

Things I've Tried:

  • Looking through the fundamental theorem of invertible matrices, and noticed that a requirement for inversion is that the columns of $A$ must be linearly independent.
  • I noticed that in a list of properties of determinants that if $A$ has a $0$ row or column, then $\det{A} = 0$. Since a row of $0$'s is not linearly independent of any other row, ever (because of the trivial sol'n),
  • I thought to myself... These seem to be conflicting statements. Can I used these two properties to show that only one must be true?

Thanks!

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Suppose $A$ is any $n \times n$ matrix.

Proof (1): Just note that $A \cdot adj(A) = \det (A)I_{n}$ from where the result follows.

Proof (2): You could also pursue something along these lines:

Suppose $\det A \neq 0$. Then the $n$ columns of $A$, namely, $C_{1}, C_{2}, ..., C_{n}$ are linearly independent. Let $X$ be a vector such that $0 = AX = x_{1}C_{1} + x_{2}C_{2} + ...+ x_{n}C_{n}$, but by linearly independence of the columns, we must have $X = 0$. But this means that $\ker A = \{0 \}$, which means that $A$ is invertible.

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