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Prove the following statement or give a counterexample if it is false

Let $M_{4}$ be the vector space of all $4$ by $4$ matrix with real entries. If $A\in M_{4}$ where rank($A$) is less than or equal to $2$, then $A$ is the subspace of $M_{4}$

What i tried

I mentioned that it is true and i do a prove by division into cases. That means i consider the cases where rank($A$)=2 , rank($A$)=1 and rank($A$)=0 and then prove each of the cases indivually.

For the case rank($A$)=0 we have only the $4$ by $4$ matrix where all entries are $0$ and clearly that is a subspace.

For the case of case of rank($A$)=1 we have the matrix of the form $$ A=\left[\begin{array}{cccc} a & b & c & d\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &0 \end{array}\right] $$

Which can be written as a linear combination of basis vectors hence a subspace.

Simillarily for rank($A$)=2 we have a matrix of the form $$ A=\left[\begin{array}{cccc} a & b & c & d\\ 0 & e & f & g\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &0 \end{array}\right] $$

Again it is a subspace since it can be written as a linear combination of basis vectors. And thus we have prove the statement for all 3 possible cases. Is my proof correct. Could anyone explain. Thanks

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  • $\begingroup$ I think the question is to check whether the set $$\{A \in M_4 : rank(A) \leq 2\}$$ is a subspace. You are not meant to work with individual matrices. $\endgroup$ – Prahlad Vaidyanathan Dec 1 '15 at 6:17
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Take $A=$

\begin{bmatrix} I_{2\times 2} && 0_{2\times 2}\\ 0_{2\times 2} && 0_{2\times 2} \end{bmatrix}

$B=$

\begin{bmatrix} 0_{2\times 2} && 0_{2\times 2}\\ 0_{2\times 2} && I_{2\times 2} \end{bmatrix}

Compute $A+B$.What is $\operatorname{Rank (A+B)}$?

$4$

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  • $\begingroup$ Yup Thanks. My issue is that for this sort of questions how would u know whether to prove or disprove. And what are the thought process involved in coming out counterexamples. It becomes easy the moment i see a counterexample but its the thought process in coming out such counterexamples by myself that i find diffcult/ $\endgroup$ – ys wong Dec 1 '15 at 6:31
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    $\begingroup$ yes you know mathematics is all about uncertainty ;what you can do is first try to just find some examples at least 3/4 then go for the proof; justify the arguements in the proof yourself and if you are still tied up consult MSE $\endgroup$ – Learnmore Dec 1 '15 at 6:35
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Let $e_{i,j}$ denote the matrix of all zeros except at entry $i,j$ where it has a $1$. Each matrix $e_{i,j}$ has rank $1$, but $e_{1,1}+e_{2,2}+e_{3,3}+e_{4,4}=I_4$ has rank $4$.

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