- #1

danago

Gold Member

- 1,122

- 4

**Let A,B be n x n matrices and e**

[tex]

\begin{array}{l}

\left\{ {x \in R^n |Ax = 2x} \right\} \\

\left\{ {x \in R^n |Ax = 2x + e_1 } \right\} \\

\left\{ {x \in R^n |Ax = Bx} \right\} \\

\left\{ {x \in R^n |Ax = Bx + e_1 } \right\} \\

\end{array}

[/tex]

_{1}be the first standard basis vector in R^{n}. For each of the following subsets of R^{n}, determine whether it is a subspace of R^{n}, giving reasons.[tex]

\begin{array}{l}

\left\{ {x \in R^n |Ax = 2x} \right\} \\

\left\{ {x \in R^n |Ax = 2x + e_1 } \right\} \\

\left\{ {x \in R^n |Ax = Bx} \right\} \\

\left\{ {x \in R^n |Ax = Bx + e_1 } \right\} \\

\end{array}

[/tex]

Now i think im ok with the second and last ones, but not so sure about the other two.

For the second and last ones, ive noticed that the zero vector does not satisfy the equation, thus the zero vector is not within the set and hence the set is not a subspace.

For the first and third, the zero vector is an element of the sets, which is one of the conditions of a subspace. The other condition would be that the set is closed under both addition and multiplication.

What i considered was to let [tex]a,b \in \left\{ {x \in R^n |Ax = 2x} \right\}[/tex], and then show that any linear combination of a and b is also within the set.

[tex]

A(ka + lb) = k(Aa) + l(Ab) = k(2a) + l(2b) = 2(ka + lb)

[/tex]

Where k and l are arbitary real constants.

And according to my understanding, this shows that any linear combination of a and b are within the set, hence the set is a subspace?

Is my reasoning correct? Id then apply the same reasoning to the third set.

Thanks,

Dan.