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Problem

$A$ is a $4 \times 4$ matrix. It is known that $\text{rank}(A)=3$. Is matrix A invertible ?

Attempt to solve

$\text{rank(A)}=3 \implies \det(A)=0$ which implies matrix is $\textbf{not}$ invertible. One dimension is lost during linear transformation if matrix is not full rank by definition. This implies determinant will be $0$ and that some information is lost in this linear transformation.

Is my intuition behind this correct ?

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  • $\begingroup$ Yes, you can even cut the lines regarding the determinant of A, and all of the rest of it still holds together. $\endgroup$ – Doug M Dec 14 '18 at 16:14
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Your intuition seems fine. How you arrive at that conclusion depends on what properties you have seen, and/or which ones you are allowed to use.

The following properties are equivalent for a square matrix $A$:

  • $A$ has full rank
  • $A$ is invertible
  • the determinant of $A$ is non-zero

There are more, but the first two are sufficient to immediately draw the desired conclusion.

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This is exactly right. You're better off mentioning the rank-nullity theorem: for a linear map $f:U \to V$ we have $$\mbox{rank}+\mbox{nullity} = \dim U$$ where the nullity is the dimension of the kernel, $\ker f$.

A four-by-four matrix represents a linear map $f: U \to V$ where $\dim U = \dim V = 4$. If the rank is three then $3+\mbox{nullity}=4$, i.e. there is a one-dimensional kernel. That means the map is not injective and has no inverse.

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$rank(A) = 3 \Rightarrow det(A) = 0$ needs to be proven, it is right though. basically, you can say that: $rank(A) < dim(A) \Rightarrow det(A) = 0$ but it still needs to be proven. an easy way to prove it is by showing that you will get a row of zeroes when trying to use raw reduction.

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If $A$ and $B$ are matrices for which $AB$ makes sense, then $$ \operatorname{rank}(AB)\le\min\{\operatorname{rank}(A),\operatorname{rank}(B)\} $$ In particular, for every $B$, $\operatorname{rank}(AB)\le\operatorname{rank}(A)=3$. Can now $AB=I$?

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