How can you show that the null space of a projection matrix is equal to the the column space of the identity matrix minus that matrix? $$N(A) = C(I – A)$$
[Math] Projection matrix and null space
linear algebramatricesprojection-matrices
Related Solutions
Let $H \in \mathbb{F}^{m \times n}$ for $\mathbb{F} = \mathbb{R}$ or $\mathbb{F} = \mathbb{C}$. If I understood you right, you want $P \in \mathbb{F}^{n \times n}$ of a maximum rank such that
- $HP = 0$,
- $P^2 = P$.
Let $H = U \Sigma V^*$ be an SVD of $H$, where $U$ and $V$ are unitary or orthogonal, depending on $\mathbb{F}$, of order $m$ and $n$, respectively, and $$\Sigma = \mathop{\rm diag}(\sigma_1, \dots, \sigma_k) \in \mathbb{R}^{m \times n}, \quad \sigma_1 \ge \cdots \ge \sigma_p > \sigma_{p+1} = \cdots \sigma_k = 0, \quad k = \min\{m,n\}.$$
Let $\Sigma' := \mathop{\rm diag}(\sigma_1', \dots, \sigma_n') \in \mathbb{R}^{n \times n}$ such that $$\sigma_i' = \begin{cases} 0, & i \le p, \\ 1, & i > p. \end{cases}.$$ Note that $\Sigma \Sigma' = 0$ and $(\Sigma')^2 = \Sigma'$.
We define $P = V \Sigma' V^*$ and check the desired properties:
- $HP = U \Sigma V^* V \Sigma' V^* = U \Sigma \Sigma' V^* = U 0 V^* = 0$,
- $P^2 = V \Sigma' V^* V \Sigma' V^* = V \Sigma' \Sigma' V^* = V (\Sigma')^2 V^* = V \Sigma' V^* = P$.
Also, if $P$ had a higher rank, than $\Sigma \Sigma' \ne 0$, so $P$ is the desired projection.
Of course, if you're doing this in Matlab (or anything else on a computer), be careful what you treat as a zero.
Part (a): By definition, the null space of the matrix $[L]$ is the space of all vectors that are sent to zero when multiplied by $[L]$. Equivalently, the null space is the set of all vectors that are sent to zero when the transformation $L$ is applied. $L$ transforms all vectors in its null space to the zero vector, no matter what transformation $L$ happens to be.
Note that in this case, our nullspace will be $V^\perp$, the orthogonal complement to $V$. Can you see why this is the case geometrically?
Part (b): In terms of transformations, the column space $L$ is the range or image of the transformation in question. In other words, the column space is the space of all possible outputs from the transformation. In our case, projecting onto $V$ will always produce a vector from $V$ and conversely, every vector in $V$ is the projection of some vector onto $V$. We conclude, then, that the column space of $[L]$ will be the entirety of the subspace $V$.
Now, what happens if we take a vector from $V$ and apply $L$ (our projection onto $V$)? Well, since the vector is in $V$, it's "already projected"; flattening it onto $V$ doesn't change it. So, for any $x$ in $V$ (which is our column space), we will find that $L(x) = x$.
Part (c): The rank is the dimension of the column space. In this case, our column space is $V$. What's it's dimension? Well, it's the span of two linearly independent vectors, so $V$ is 2-dimensional. So, the rank of $[L]$ is $2$.
We know that the nullity is $V^\perp$. Since $V$ has dimension $2$ in the $4$-dimensional $\Bbb R^4$, $V^\perp$ will have dimension $4 - 2 = 2$. So, the nullity of $[L]$ is $2$.
Alternatively, it was enough to know the rank: the rank-nullity theorem tells us that since the dimension of the overall (starting) space is $4$ and the rank is $2$, the nullity must be $4 - 2 = 2$.
Best Answer
The column space of a matrix is the same as the image of the transformation. (that's not very difficult to see but if you don't see it post a comment and I can give a proof)
Now for $v\in N(A)$, $Av=0$ Then $(I-A)v=Iv-Av=v-0=v$ hence $v$ is the image of $I-A$.
On the other hand if $v$ is the image of $I-A$, $v=(I-A)w$ for some vector $w$. Then $$ Av=A(I-A)w=Aw-A^2w=Aw-Aw=0 $$ where I used the fact $A^2=A$ ($A$ is projection). Then $v\in N(A)$.