Which of the following sets of $n \times n$ matrices with real entries is a vector space over $\mathbb R$ if the vector addition and scalar multiplication are as usual?

A) the set of invertible matrices,

B) the set of non-invertible matrices,

C) the set of matrices with a zero diagonal,

D) the set of non-zero matrices,

E) none of the above.

I have the answer. It is C. I just do not understand why. I thought it must be A because of the axiom that says there must be an inverse of each element. Why the zero diagonal?


Choice A is not since $I$ and $-I$ are invertible but $I + (-I) = \textbf{0}$ is not invertible.

Choice B doesn't work since the matrices

$$ \left [ \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array} \right ] + \left [ \begin{array}{cc} 0 & 0 \\ 0 & 1 \\ \end{array} \right ] \;\; =\;\; I $$

But neither of the original matrices is invertible while their sum is.

Choice C is correct since adding any two matrices with zeros along the diagonal maintain zeros along the diagonal. The same is true if we multiply one of these matrices by any scalar.

Choice D also is false since $I + (-I) = \textbf{0}$ again.

  • $\begingroup$ Oh thank you. I see now. So C is the only one where if you use vector addition or scalar multiplication, you get another element that will always be in the same set. $\endgroup$ – StephanCasey Mar 18 '15 at 7:45
  • 1
    $\begingroup$ @StephanCasey Exactly. In a vector space you need the sum of elements or scalar multiples of elements to keep the same properties by which you defined the vector space. If it is not closed under these for a single instance, then it's not a vector space. $\endgroup$ – Mnifldz Mar 18 '15 at 7:50

Be careful: you can have an inverse for addition but not for multiplication - or vice versa. Which one is a part of the vector space axioms?


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