Intereting Posts

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How to solve this to find the Null Space
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Product of two probability kernel is a probability kernel?
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Extensions of degree two are Galois Extensions.
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Conjugacy classes of D2n?
Question on Good Pairs
More than 99% of groups of order less than 2000 are of order 1024?
probability question (“birthday paradox”)
If $a > b$, is $a^2 > b^2$?

Seems like I still don’t get it, I think I am missing something important.

Let $V$ be an $n$ dimensional inner product space ($n \geq 1$), and $T\colon\mathbf{V}\to\mathbf{V}$ be a linear transformation such that:

- $T^2 = T$
- $||T(a)|| \leq ||a||$ for every vector $a$ in $\mathbf{V}$;
Prove that a subspace $U \subseteq V$ exist, such that $T$ is the orthogonal projection on $U$.

- Norms Induced by Inner Products and the Parallelogram Law
- Span of an empty set is the zero vector
- Does the rank-nullity theorem hold for infinite dimensional $V$?
- If a field $F$ is such that $\left|F\right|>n-1$ why is $V$ a vector space over $F$ not equal to the union of proper subspaces of $V$
- Cross product as result of projections
- Determine if vectors are linearly independent

Now, I know these things:

- The fact hat $T^2 = T$ guarantees that $T$ is indeed a projection, so I need to prove that T is an
**orthogonal**projection (I guess this is where $||T(a)|| \leq ||a||$ kicks in). - To do this I can prove that:
- For every $v$ in $ImT^{\perp}$, $T(v) = 0$
- Alternatively, I can prove that for every $v$ in $ImT$ and $u$ in $KerT$, $(v,u)=0$.
- $T$ is self-adjoint (according to Wikipedia)
- The matrix $A = [T]_{E}$ when $E$ is an orthonormal basis, is hermitian (this is equivalent to the previous point).
- What else?

I’ve been thinking about it for quite some time now, and I’m pretty sure there is something big I’m missing, **again**. I just don’t know how to use the data to prove any of these things.

Thanks!

- Calculating the determinant gives $(a^2+b^2+c^2+d^2)^2$?
- Show determinant of matrix is non-zero
- Solutions to the matrix equation $\mathbf{AB-BA=I}$ over general fields
- How can I determine if three 3d vectors are creating a triangle
- Is the identity matrix the only matrix which is its own inverse?
- What is the standard proof that $\dim(k^{\mathbb N})$ is uncountable?
- Mnemonics for linear algebra
- Circulant determinants
- Determinant of rank-one perturbation of a diagonal matrix
- Is function invertible?

One approach to this is by using Pythagorean Theorem. You fill in the details.

Indeed, suppose that $T \psi \in Im T$. Note that $V = Ker T \oplus Ker T^\bot$. Now write $T \psi = k + k'$ where $k \in Ker T$ and $k' \in Ker T^\bot$. We get that

$$\|k'\|^2 \ge \|T k'\|^2 = \|T k + T k'\|^2 = \|T(T \psi)\|^2 = \|T \psi\|^2 = \|k\|^2 + \|k'^2\|,$$

from which we gather that $k = 0$. Therefore $T \psi \in Ker T^\bot$.

Thus $Im T \subset Ker T^\bot$. On the other hand if $\phi \in Im T^\bot \cap Ker T^\bot$, then one can show that $\phi \in Im T^* \cap Ker T^*$. (Because $Im T^\bot = Ker T^*$.) One can also show that $T^*$ is a projection. Let $\phi = T^* \psi$. Then $T^* \psi = T^* T^* \psi = T^* \phi = 0$. Hence $\phi = 0$. Since $V = Im T \oplus Im T^\bot$, we get that $Ker T^\bot \subset Im T$.

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