1. ## Linear Spaces

1) If v is any vector in R^n, what are the possible dimensions of the space of all n x n matrices A such that Av = 0?

2) Find a basis of each of the following linear spaces, and thus determine their dimensions.
a) {f in P4: f is even}
b) {f in P4: f is odd}

2. Generally speaking, one equation reduces the number of "free variables", and so the dimension, by 1. When you multiply a matrix, A, by the fixed vector, v, and set it equal to 0: Av= 0, each row of the matrix is one equation. Thus while the dimension of the space of all n by n matrices is $n^2$ the fact that Av= 0 gives n equations, reducing the dimension to $n^2- n= n(n-1)$.

P4 is the set of polynomials of degree 4 or less? Then your first space consists of all such polynomials that have only even powers: $\{ax^4+ bx^2+ c\}$. The second is all such polynomials that have only odd powers: $\{ax^3+ bx\}$. It should be easy to find bases for those.

3. Originally Posted by HallsofIvy
Generally speaking, one equation reduces the number of "free variables", and so the dimension, by 1. When you multiply a matrix, A, by the fixed vector, v, and set it equal to 0: Av= 0, each row of the matrix is one equation. Thus while the dimension of the space of all n by n matrices is $n^2$ the fact that Av= 0 gives n equations, reducing the dimension to [tex]n^2- n= n(n-1)[tex].

@HallsofIvy - Would you please explaining this - or suggesting somewhere from where I can read it up? Question I have is
" one equation reduces the number of "free variables", and so the dimension, by 1" - By this I guess you mean the dimension of the null-space of the equations. How are you relating that to dimension of the matrix space? " $n^2- n= n(n-1)$."
Thanks

4. For example. Find the dimension of the subspace of $R^4$, <u, v, w, x>, satisfying 3u- 2v+ 3w- x= 0. (Yes, we can consider it the null space of the operator that maps any <u, v, w, x> to 3u- 2v+ 3w- x in $R^1$.) Since that is a single equation I know immediately that the dimension of this subspace is 4- 1= 3. In fact, I can solve for x, x= 3u- 2v+ 3w, so I can write any member of that subspace as <u, v, w, 3u- 2v+3w>= u(1, 0, 0, 3>+ v<0, 1, 0, -2>+ w<0, 0, 1, 3> showing that {<1, 0, 0, 3>, <0, 1, 0, -2>, <0, 0, 1, 3>} is a basis.

If I were talking about the subspace satifying both that and the equation u+ v- w+ x= 0 (Which we can think of as the null space for the linear transformation that takes <u, v, w, x> to <3u-2v+3w-x. u+v-w+x> in $R^2$. Any subspace is the null space of some linear transformation!), I can see that there are two equations and, once I have checked that they are independent, know that this subspace will have dimension 4- 2= 2. If I add the two equations, I eliminate x and get 4u- v+ 2w= 0. I can solve that for v: v= 4u+ 2w. Putting that into the first equation, 3u-2(4u+2w)+ 3w- x= -5u- w- x= 0 so x= -5u- w. That means that any member of this subspace can be written as <u, 4u+2w, w, -5u- w>= u<1, 4, 0, -5>+ v<0, 2, 1, -1> which means that {<1, 4, 0, -5>,<0, 2, 1, -1>} is a basis for the subspace.

5. Originally Posted by HallsofIvy
For example. Find the dimension of the subspace of R4, <u, v, w, x>, satisfying 3u- 2v+ 3w- x= 0. (Yes, we can consider it the null space of the operator that maps any <u, v, w, x> to 3u- 2v+ 3w- x in $R^1$.) Since that is a single equation I know immediately that the dimension of this subspace is 4- 1= 3. In fact, I can solve for x, x= 3u- 2v+ 3w, so I can write any member of that subspace as <u, v, w, 3u- 2v+3w>= u(1, 0, 0, 3>+ v<0, 1, 0, -2>+ w<0, 0, 1, 3> showing that {<1, 0, 0, 3>, <0, 1, 0, -2>, <0, 0, 1, 3>} is a basis.

If I were talking about the subspace satifying both that and the equation u+ v- w+ x= 0 (Which we can think of as the null space for the linear transformation that takes <u, v, w, x> to <3u-2v+3w-x. u+v-w+x> in $R^2$. Any subspace is the null space of some linear transformation!), I can see that there are two equations and, once I have checked that they are independent, know that this subspace will have dimension 4- 2= 2. If I add the two equations, I eliminate x and get 4u- v+ 2w= 0. I can solve that for v: v= 4u+ 2w. Putting that into the first equation, 3u-2(4u+2w)+ 3w- x= -5u- w- x= 0 so x= -5u- w. That means that any member of this subspace can be written as <u, 4u+2w, w, -5u- w>= u<1, 4, 0, -5>+ v<0, 2, 1, -1> which means that {<1, 4, 0, -5>,<0, 2, 1, -1>} is a basis for the subspace.
Hi - Thanks for a detailed reply. I will read it carefully and post any further questions plz.