How Do You Prove the Dimension of a Direct Sum Equals the Sum of Dimensions?

[jecharla]

Exercise #17 in Linear Algebra done right is to prove that the dimension of the direct sum of subspaces of V is equal to the sum of the dimensions of the individual subspaces. I have been trying to figure this out for a few days now and I'm really stuck. Here's what I have got so far:

Choose a base of each of the subspaces and combine them in one list which clearly spans V and has the length we are looking for. Now we just have to show that this list is linearly independent.

If this list is not linearly independent than one of the vectors is in the span of the previous vectors. If I could show that this vector must be in the span of one of the other bases, then that would prove the result but I can't quite figure out how to get there. Any help would be greatly appreciated.

 

[DonAntonio]

Exercise #17 in Linear Algebra done right is to prove that the dimension of the direct sum of subspaces of V is equal to the sum of the dimensions of the individual subspaces. I have been trying to figure this out for a few days now and I'm really stuck. Here's what I have got so far:

Choose a base of each of the subspaces and combine them in one list which clearly spans V and has the length we are looking for. Now we just have to show that this list is linearly independent.

If this list is not linearly independent than one of the vectors is in the span of the previous vectors. If I could show that this vector must be in the span of one of the other bases, then that would prove the result but I can't quite figure out how to get there. Any help would be greatly appreciated.

Let's do it for two subspaces as this summarizes the logical argument: suppose A:=\{v_1,...,v_k\}\,\,,\,\,B:=\{w_1,...,w_r\}\, are basis resp. of subspaces \,U,W\leq V\,\,,\,\,with\,\,\,U\cap W=\{0\}.

Suppose one of the vectors in \,\{v_1,...,v_k,w_1,...,w_r\}\, depends linearly on the preceeding ones. Then this must be

one of the \,w_i\,'s (why?) , so \,w_i\in Span\{v_1,...,v_k,w_1,...,w_{i-1}\}\,\Longrightarrow w_i=a_1v_1+...+a_kv_k+b_1w_1+...+b_{i-1}w_{i-1} We then get a_1v_1+...+a_kv_k=-b_1w_1-...-b_{i-1}w_{i-1}+w_i\in V\cap W=\{0\}

From here, \,a_1v_1+...+a_kv_k=0=-b_1w_1-...-b_{i-1}w_{i-1}+w_i\, , from where we reach at once our contradiction.

DonAntonio

 

[HallsofIvy]

Exercise #17 in Linear Algebra done right is to prove that the dimension of the direct sum of subspaces of V is equal to the sum of the dimensions of the individual subspaces.

As stated, you can't prove this, it isn't true. It is true and you can prove it if you add the condition that the only vector contained in both subspaces is the 0 vector. Of course, your proof will have to use that condition.

I have been trying to figure this out for a few days now and I'm really stuck. Here's what I have got so far:

Choose a base of each of the subspaces and combine them in one list which clearly spans V and has the length we are looking for. Now we just have to show that this list is linearly independent.

If this list is not linearly independent than one of the vectors is in the span of the previous vectors. If I could show that this vector must be in the span of one of the other bases, then that would prove the result but I can't quite figure out how to get there. Any help would be greatly appreciated.

If they were not you could construct a non-zero vector that is in both subspaces.

 

[DonAntonio]

As stated, you can't prove this, it isn't true. It is true and you can prove it if you add the condition that the only vector contained in both subspaces is the 0 vector. Of course, your proof will have to use that condition.

This condition is implicit in the direct sum thing: cero must be the intersection of any space and the sum of all the others, otherwise the sum is not direct.

Donantonio

If they were not you could construct a non-zero vector that is in both subspaces.