Help with Fundamental Theorem of Homomorphisms

**sashikanth** · March 19th 2012, 09:20 AM

Let $V$ be the set of vectors in $\ \Re^4$ of the form $(x, x+y, x-y, y)$ where $x, y \ \epsilon \ \Re$ . Find a subspace $W \ \epsilon \ \Re^4$ such that $\ \Re^4 = V \oplus W$ .

Attempt at a solution:

Now it is easy to prove that $V$ is of dimension 2 with a basis as $[(1,1,-1,0), \ (0,1,1,1)]$ . All we need is a 2 dimensional subspace $W$ such that the basis of both $V \mbox{and} \ W$ have no element in common.

By trial, $W$ with a basis as $[(1,1,1,1), \ (1,0,1,0)]$ satisfies the given condition. My question is, is there a systematic method to solve such questions? What do I do if the basis of the involved vector spaces are so high that trial and error don't work?

Thanks!

**MATHNEM** · March 19th 2012, 10:55 AM

Originally Posted by sashikanth

By trial, $W$ with a basis as $[(1,1,1,1), \ (1,0,1,0)]$ satisfies the given condition. My question is, is there a systematic method to solve such questions? What do I do if the basis of the involved vector spaces are so high that trial and error don't work?

One possible approach is to take a basis $S$ for $\Re^4$ , such as the canonical one, and look for those vectors in $S$ which are not linear combination of the ones in a basis for $V$ . In the worst case (for your problem), you'll have to do this for three vectors of $S$ .

I'm not sure whether this is the way you found that basis for $W$ .

**sashikanth** · March 19th 2012, 11:03 AM

Originally Posted by MATHNEM

One possible approach is to take a basis $S$ for $\Re^4$ , such as the canonical one, and look for those vectors in $S$ which are not linear combination of the ones in a basis for $V$ . In the worst case (for your problem), you'll have to do this for three vectors of $S$ .

I'm not sure whether this is the way you found that basis for $W$ .

I did the exact reverse: I extended the basis of $V$ to firstly, a set with 3 Linearly Independent Vectors, and then I extended that set to one containing 4 linearly independent vectors. The 2 new vectors added constitute the basis of W.

I'm guessing that this approach of trial might be the only way to solve such problems. I mean, there wouldn't be any generic method to GENERATE a vector that is NOT a linear combination of given vectors, isn't it?

**HallsofIvy** · March 19th 2012, 05:56 PM

Originally Posted by sashikanth

Let $V$ be the set of vectors in $\ \Re^4$ of the form $(x, x+y, x-y, y)$ where $x, y \ \epsilon \ \Re$ . Find a subspace $W \ \epsilon \ \Re^4$ such that $\ \Re^4 = V \oplus W$ .

Attempt at a solution:

Now it is easy to prove that $V$ is of dimension 2 with a basis as $[(1,1,-1,0), \ (0,1,1,1)]$ .

No, V is of dimension 2 with basis (1, 1, 1, 0) and (0, 1, -1, 1). (1, 1, -1, 0) and (0, 1, 1, 1) are not even in V.

All we need is a 2 dimensional subspace $W$ such that the basis of both $V \mbox{and} \ W$ have no element in common.

A vector that is clearly normal to (1, 1, 1, 0) and (0, 1, -1, 1) is (-1, 0, 1, 1). Another is (-1, 1, 0, -1). Since those are independent, they form a basis for W.

By trial, $W$ with a basis as $[(1,1,1,1), \ (1,0,1,0)]$ satisfies the given condition. My question is, is there a systematic method to solve such questions? What do I do if the basis of the involved vector spaces are so high that trial and error don't work?

Thanks!

**sashikanth** · March 19th 2012, 07:28 PM

Originally Posted by HallsofIvy

No, V is of dimension 2 with basis (1, 1, 1, 0) and (0, 1, -1, 1). (1, 1, -1, 0) and (0, 1, 1, 1) are not even in V.

A vector that is clearly normal to (1, 1, 1, 0) and (0, 1, -1, 1) is (-1, 0, 1, 1). Another is (-1, 1, 0, -1). Since those are independent, they form a basis for W.

Sorry my mistake. The actual question was $(x, x+y, y-x, y)$ . I made a typo above. My question still remains: you have obtained the 2 vectors $(-1,0,1,1), \ (-1,1,0,-1)$ in essentially the same way as I have (outlined in the above post). Is there any way to do it without trial and error? In other words, suppose $V$ is a 7 dimensional subspace of $\Re^{10}$ and you need to find a suitable $W$ , how would one go about solving the problem? Obtaining the 2 vectors like how we have in this case by observation would be really tough

**Deveno** · March 20th 2012, 08:48 PM

note that (x,x+y,y-x,y) = (x,x,-x,0) + (0,y,y,y) = x(1,1,-1,0) + y(0,1,1,1). thus {(1,1,-1,0),(0,1,1,1)} is clearly a spanning set. a formal proof that it is LI (linearly independent) is not hard:

suppose a(1,1,-1,0) + b(0,1,1,1) = (0,0,0,0).

that is: (a,a+b,b-a,b) = (0,0,0,0). from the first coordinate, we see that a = 0, and from the fourth we see that b = 0.

so we have a basis. so how to find "two more" LI vectors in R⁴, so that we can extend a basis of W to a basis of R⁴?

one way that works, if you have an inner product, is to consider the orthogonal complement of V:

{(x,y,z,w) in R⁴: (x,y,z,w).(a(1,1,-1,0) + b(0,1,1,1)) = 0}.

this gives "a ugly formula" xa + y(a+b) + z(b-a) + wb = 0, but...we are free to choose convenient values for a and b to help us:

so let's use a = 1, b = 0, and see what we get:

x + y - z = 0 (*)

and for our second "test value" let's choose a = 0, b = 1:

y + z + w = 0 (**)

this is not so bad, note that choosing x and y, then determines z and w. can you see why x = y = 0 is a "bad" choice? can you also see that choosing (x,y) = (1,0) and (x,y) = (0,1) leads to 2 linearly independent vectors (x,y,z,w)?

the first choice leads to the vector (1,0,1,-1), and the second choice leads to the vector (0,1,1,-2).

it is fairly straight-forward that both of these vectors are orthogonal to V:

(1,1,-1,0).(1,0,1,-1) = 1 + 0 - 1 + 0 = 0
(0,1,1,1).(1,0,1,-1) = 0 + 0 + 1 - 1 = 0

(1,1,-1,0).(0,1,1,-2) = 0 + 1 - 1 + 0 = 0
(0,1,1,1).(0,1,1,-2) = 0 + 1 + 1 - 2 = 0

it is also straight-forward to see that these two vectors are not in V:

the third coordinate is not the additive inverse (negative) of the second, as is the case with every element of V.

as we saw above, these two vectors are LI, so the subspace they span is of dimension 2.

so we have 2 subspaces of R⁴, V and W. so, how can we tell if R⁴ = V⊕W?

recall that dim(V+W) = dim(V) + dim(W) - dim(V∩W), and that V+W = V⊕W if and only if dim(V∩W) = 0,

which happens if and only V∩W = {0}.

so here, we have dim(V+W) = 2 + 2 - dim(V∩W).

but we know that every vector in W is orthogonal to every vector in V. so if w is in W, and w is in V as well,

it must be that w is orthogonal to itself: w.w = 0. the left side is a sum of squares, so every coordinate of w is 0, w = (0,0,0,0).

so V∩W = {(0,0,0,0)}, and so dim(V∩W) = 0, hence V+W = V⊕W, and has dimension 2 + 2 - 0 = 4.

we can go further, and write W solely in terms of z and w:

W = {(z,w,z+w,-z-2w): z,w in R}, and we can write any vector in R⁴ uniquely as:

x(1,1,-1,0) + y(0,1,1,1) + z(1,0,1,-1) + w(0,1,1,-2).

we can put this another way: the solution of the system of linear equations,

x + z + w = a
x + y + w = b
-x + y + z + w = c
y - z - 2w = d or the matrix equation:

$\begin{bmatrix}1&0&1&1\\1&1&0&1\\-1&1&1&1\\0&1&-1&-2 \end{bmatrix} \begin{bmatrix}x\\y\\z\\w \end{bmatrix} = \begin{bmatrix}a\\b\\c\\d \end{bmatrix}$

has a unique solution.

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