LINEAR ALGEBRA: Consider 2X2 Matrices - What are the subspaces?

[VinnyCee]

Consider 2-by-2 matrices \mathbf{A} =\left( \begin{array}{cc}a & b \\c & d \\\end{array} \right) \in \mathbbm{R}^{2 X 2}. Which of the following are subspaces of \mathbbm{R}^{2 X 2}?

(A) {A | c = 0}

(B) {A | a + d = 0}

(C) {A | ad - bc = 0}

(D) {A | b = c}

(E) {A | Av = 2v}, where v is some vector in R^2.






I think that all except the last would be subspaces of the 2 X 2 set of matrices. I know that choice (C) is the determinent of the matrix, which is in the subspace of it's own matrix, right?

Also, how would I make a case for the first four choices using the 3 "rules" that determine if a vector is a subspace of another?

(1) {0} must be in the subset for it to be a subset.

(2) if v is in W then a * v is in W for all real numbers a.

(3) if u and v are in W, then u + v is in W.

 

[Daverz]

For (C), you should be able to come up with 2 matrices with zero determinant that add to a matrix with non-zero determinant.

The set (E) obviously violates one of the rules.

 

[radou]

(C) {A | ad - bc = 0}

Let
\mathbf{U} =\left( \begin{array}{cc}a & b \\c & d \\\end{array} \right) \in S and
\mathbf{V} =\left( \begin{array}{cc}e & f \\g & h \\\end{array} \right) \in S, where S is the subspace of \mathbbm{R}^{2 X 2} described with (C).
Then, U + V = W has to be in S.
\mathbf{W} =\left( \begin{array}{cc}a+e & b+f \\c+g & d+h \\\end{array} \right) \in S. Further on, (a+e)(d+h)-(c+g)(b+f)=0 implies \frac{ah+ed}{bg-fc}=1, from which we obtain bg \neq fc (1). So, we have shown that W is in the same subspace as U and V just for the matrices U and V such that (1) is true. But, it should be true for every two matrices U and V satisfying only the property stated in (C). Hence, (C) is no subspace.

 

[VinnyCee]

So, in order to do this "proving" stuff. A person should assign variables (presumably the set of all real numbers) to two of the types of spaces (or matrix m X n, etc.) being questioned as subspaces? Then apply the three rules to see if they are broken?

How, exactly, do we know that the U and V matrices above do not satisfy rule number 3(if u and v are in W, then u + v is in W)?

 

[Daverz]

Radou did a proof by contradiction. Assume what you want to prove and show that it leads to a contradiction.

The one I suggested would be a proof by counterexample.

 

[VinnyCee]

Only (A), (B) and (D) are subspaces?

 

[Daverz]

Yeah, you need to show that. Radou basically gave you the template for rule 3: show that for arbitrary U and V in the subset, W = U + V is also in the subset.

 

[VinnyCee]

For example, I can use rule 3 for case (A) like so?:

\mathbf{U} =\left( \begin{array}{cc}a & b \\0 & d \\\end{array} \right) \in S

\mathbf{V} =\left( \begin{array}{cc}e & f \\0 & h \\\end{array} \right) \in S

\mathbf{W} =\left( \begin{array}{cc}a+e & b+f \\0+0 & d+h \\\end{array} \right) \in S


Does this prove (for rule 3 only) that case (A) is a subspace of R^{2 X 2}?

EDIT:

For rule 2 -

a\,\left( \begin{array}{cc}a & b \\0 & d \\\end{array} \right) \in S, where the first a is a real number.

is true?

 

[VinnyCee]

"If v \in W then av \in W for all a \in R."

This is rule 2, how would I go about defineing a vector v that is in W?

\overrightarrow{v}\,=\,v_1,\,v_2,\,...,\,v_n?

 

[Daverz]

For example, I can use rule 3 for case (A) like so?:

\mathbf{U} =\left( \begin{array}{cc}a & b \\0 & d \\\end{array} \right) \in S

\mathbf{V} =\left( \begin{array}{cc}e & f \\0 & h \\\end{array} \right) \in S

\mathbf{W} =\left( \begin{array}{cc}a+e & b+f \\0+0 & d+h \\\end{array} \right) \in SDoes this prove (for rule 3 only) that case (A) is a subspace of R^{2 X 2}?

Well, simplify everything that can be simplified, and it should be obvious that you have a new element that's a member of the subset.

EDIT:

For rule 2 -

a\,\left( \begin{array}{cc}a & b \\0 & d \\\end{array} \right) \in S, where the first a is a real number.

is true?

Well, choose a variable name that's not going to confuse the real number multiplier with one of the matrix elements.

Then...do the matrix arithmetic. That's basically all you do to check rules (2) & (3): do the matrix arithmetic for the particular subspace rule, simplify whatever can be simplified, then check to see if the resulting matrix is a member of the subset according to the definition (which, if it's not obvious by inspection, is just more matrix arithmetic). For rule (1), you just need to check that the 0 matrix is a legal member of the subset.

 

[Daverz]

"If v \in W then av \in W for all a \in R."

This is rule 2, how would I go about defineing a vector v that is in W?

\overrightarrow{v}\,=\,v_1,\,v_2,\,...,\,v_n?

The vectors in this space are 2x2 matrices, so just construct an arbitrary 2x2 matrix like V above.