More vector space questions

[jsbeckton]

Have a vector space problem that has me confused, it reads as follows:

Let \(\displaystyle S = \{ A \in M_2 (R):\det (A) = 0\}\)

a) Is the Zero vector from \(\displaystyle M_2 (R)\) in S?
A 2x2 0 matrix would have a determinate of 0 so I believe that it is in S.

b) Give an explicit example illistrating that S is not closed under matrix addition.
Any example that has real numbers and does not have a determinate of 0 will due, eg:

\(\displaystyle \matrix{
1 & 2 \cr
3 & 4 \cr\) ....... has a det(A) = -2 ....... not 0

c) Is S closed under scalar multiplication? Justify your anwser.
I believe that it is not closed under scalar multiplication since multiplying the matrix and the determinate by a scalar will not hold up but am not sure about that or how to justify it. Any help would be greatly appreciated. Thanks!

 

[pka]

About using TeXaide.

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_________________________________________
You are correct about a)

For b) you will need two examples: these matrices are in S
\(\displaystyle \left[ {\begin{array}
1 & 1 \\
0 & 0 \\
\end{array}} \right]{\rm{ and }}\left[ {\begin{array}
0 & 0 \\
2 & 1 \\
\end{array}} \right]\). Both have det=0 but what about the sum?
If \(\displaystyle \det (A) = 0{\rm{ then }}\det (\alpha A) = 0\).
Therefore you need to rethink part c).

 

[jsbeckton]

\(\displaystyle \[
a^2 + b^2 = c^2
\]\)

just trying something, thanks for the tips.

 

[jsbeckton]

b) A + B does not have a determinate of zero so A+B is not an element of S, S is not closed.

c) Im thinking that since the det(A)= 0 and det(B) = 0 and the determinate of 5A or 5B also will be 0, then it is closed under scalar multiplication. Am I correct?

 

[stapel]

You're posting some cool questions....

a) Yes, the zero matrix is in S.

b) If a matrix (such as in your example) does not have a determinant (note spelling) of zero, then it isn't in S, so it isn't fair game for a start of the exercise. You need to pick two matrices, as demonstrated by pka, that are in S (that is, two matrices with determinants of zero), such that their sum matrix has a non-zero determinant.

c) Not sure what you mean, in your first post, by "not holding up". You should have some result, as hinted to by pka, that says "det(kA) = k×det(A)". So if det(A) = 0, what can you say about det(kA)? So is kA in S? (You don't need to consider any other matrix "B", since you're only multiplying a scalar on one matrix, not adding two matrices.)

Eliz.

 

[jsbeckton]

Thanks for the help

det(kA) = k×det(A)". So if det(A) = 0

therefore the scalar will always be multiplied by 0 since the det(A) always equals 0. s is closed under scalar multiplication.