Finding the projections associated to the direct sum decomposition of a vector space

Question

PROBLEM STATEMENT

Hi all. First, I am given two vector spaces:

Let $V = \mathbb{R}^4$. Consider the following subspaces: $V_1 = \{(x,y,z,t)\ : x = y = z\}, V_2=[(2,1,1,1)], V_3 =[(2,2,1,1)]$

And let $V = M_n(\mathbb{k})$. Consider the following subspaces:

$V_1 = \{(a_{ij}) \in V : a_{ij} = 0,\forall i < j\}$

$V_2 = \{(a_{ij}) \in V : a_{ij} = 0,\forall i > j\}$

$V_3 = \{(a_{ij}) \in V: a_{ij} = 0, \forall i \neq j\}$

Prove that for both statements it is true that $V = V_1 \oplus V_2 \oplus V_3$

Find the projections associated to both decompositions.

ATTEMPT AT A SOLUTION

So for both it was fairly trivial to prove the direct sum. To be explicit about my method, in the first case I took the basis of $V_1$ to be $\{(1,0,0,0),(0,0,0,1)\}$ (from $\{(1,1,1,0),(0,0,0,1)\}$ given by the definition of the subspace) and for $V_2, V_3$ the given vectors. I used these as columns, performed matrix operations to reach row echelon form, and thereby discovered that I could form a diagonal matrix and that each column was linearly independent.

From there, the sum clearly gave me a general vector in $V$, and by the definition of their linear independence the intersection between the spaces is necessarily $\{0\}$. Therefore we have a direct sum in the first case.

In the second case, the basis of $V_1$ is a matrix with entries of $1$ above the diagonal and $0$ along and below the diagonal, $V_2$ is $0$ along and above the diagonal and $1$ below, and the basis of $V_3$ is a diagonal matrix with $0$ above and below.

Again the direct sum is clear.

HELP NEEDED

However, I do not know how to find the projections associated to these decompositions. I do know the definition of a a projection. Is there some intuitive way to see what it would be like here or in similar cases, or do I need to apply the formula $P_x = A(A^TA)^{-1}A^T$? If so, could I see an example from the first case?

Answer 1

Consider, for instance the projection of $\mathbb{R}^4$ on $V_1$. Since $\bigl((1,0,0,0),(0,0,0,1)\bigr)$ is a basis of $V_1$, $\bigl((2,1,1,1)\bigr)$ is a basis of $V_2$, and $\bigl(2,2,1,1)\bigr)$ is a basis of $V_3$, you consider the linear map $P\colon\mathbb{R}^4\longrightarrow\mathbb{R}^4$ such that

• $P(1,0,0,0)=(1,0,0,0)$;
• $P(0,0,0,1)=(0,0,0,1)$;
• $P(2,1,1,1)=(0,0,0,0)$;
• $P(2,2,1,1)=(0,0,0,0)$.

This is the projection of $\mathbb{R}^4$ onto $V_1$ based upon the decomposition $\mathbb{R}^4=V_1\oplus V_2\oplus V_3$.