1.) The set W of all 2x3 matrices of the form

a b c

a 0 0

where c = a + b, is a subspace of M23 (Matrics 23). Show that every vector in W is a linear combination of

 

W1 =

1 0 1

1 0 0

 

W2 =

0 1 1

0 0 0

 

Do I have to combine both W1 and W2 into one equation?

Edited March 5, 201114 yr by hkus10

Yes, that'd work. You can start with:

 

[math]a \begin{pmatrix}1 & 0 & 1\\ 1 & 0 & 0\end{pmatrix} + b \begin{pmatrix}0 & 1 & 1\\ 0 & 0 & 0\end{pmatrix}[/math]

 

and see where that leads you.

Yes, that'd work. You can start with:

 

[math]a \begin{pmatrix}1 & 0 & 1\\ 1 & 0 & 0\end{pmatrix} + b \begin{pmatrix}0 & 1 & 1\\ 0 & 0 & 0\end{pmatrix}[/math]

 

and see where that leads you.

 

 

What I get is

[math]\begin{pmatrix}a & b & 2c\\ a & 0 & 0\end{pmatrix}[/math]

 

which is not

[math]\begin{pmatrix}a & b & c\\ a & 0 & 0\end{pmatrix}[/math]

Edited March 6, 201114 yr by hkus10

You've made an error in that upper-right matrix element, then.

You've made an error in that upper-right matrix element, then.

Is this the answer? aW_1 + bW_2 = [math] \begin{pmatrix}a & b & a+b\\ a & 0 & 0\end{pmatrix} [/math] where a, b can be any real number.

Edited March 6, 201114 yr by hkus10

What I get is

[math]\begin{pmatrix}a & b & 2c\\ a & 0 & 0\end{pmatrix}[/math]

 

How did you get an expression involving c starting from an expression involving only a and b ?

 

 

You cannot do that without some other expression relating a and b to c.

The problem stipulates "where c = a + b".

 

hkus10: Yes. And a + b = c, so the upper-right cell is equivalent to c. Not 2c.