Which of these linear algebra statements are true?

Question

let V= R^5 be equipped with the usual euclidean inner-product. Which of the following statements are true?

a). If W and Z are subspaces of V such that both of them are of dimension 3, then there exists z ∈ Z such that z is not equal to 0 and z ⊥ W.

b). There exists a non-zero linear map T : V → V such that ker(T) ∩ W not equal to {0} for every subspace W of V of dimension 4.

c). Let W be a subspace of V of dimension 3. Let T : V →W be a linear map which is surjective and let S : W → V be a linear map which is injective Then, there exists x ∈ V such that x is not equal 0 such that S ◦ T(x) = 0.

My attempts: for option a) i take dim V=5 and dim w=3,dimz=3,,and i know that Dim(w+z)=5 =DimV....Dim(w+z)= dimw + dimz - dim w ∩ dimz = 3+3-0 =6 and i know that dim w ∩ dimz={0} because z ⊥ W. but dim(z+w) =5 not equal to 6..so option a) is incorrect.

for option b) ker(T)∩ W is not equal {0},,that mean kernel(T) =1 so by rank nullity theorem dim W of V =4 as we know that DimV = 5 and dim null(w) =1 so we get Dimrange(W) =4 so the option B is correct.

for option C) i don't know...i have no any hint to solve this ...

if anbody try to rectified my mistake i would be very thankful..

Answer 1

a) False. Take any two equal spaces $W$ and $Z$.

b) True. Let $T$ be such that $\operatorname{rank}T=1$. Then $\dim\ker T=4$ and, in a $5$-dimensional space, any two $2$-dimensional subspaces have non-trivial intersection.

c) True. $\operatorname{rank}T\leqslant 3$ and therefore $\operatorname{rank}(S\circ T)\leqslant 3$. Since $\dim V=5$, $\det(S\circ T)=0$ and therefore $\ker(S\circ T)\neq\{0\}$.

Answer 2

a) Unachieveable for W=Z, because any $z \in Z$ will be orthogonal to itself $z \in W$.

b) If T is non-zero, its kernel dimension is at most 4. In 5D space any two 4D subspaces will have shared dimensions, and hence they will have intersections along basis vectors of that dimension. So it is true

c) Kernel of an injective function is $\{0\}$, so the output of $T$ can not be 0. But $T$ can map to the kernel of $S$. If the kernel of $S$ is at least 3D, then it will intersect with the output of $T$. The kernel of $S$ is 2D, and it is exactly disjoint with $W$, since $S:W\rightarrow V$, so it is not possible

Answer 3

Your attempt for a):

i take dim V=5 and dim w=3,dimz=3,,and i know that Dim(w+z)=5 =DimV....Dim(w+z)= dimw + dimz - dim w ∩ dimz = 3+3-0 =6 and i know that dim w ∩ dimz={0} because z ⊥ W. but dim(z+w) =5 not equal to 6..so option a) is incorrect.

It is not clear which question you're trying to answer, but your statement seems completely irrelevant. My only guess as to your thought process is that you were trying to fit in the formula $\dim (W+Z)= \dim W + \dim Z - \dim (W ∩ Z)$ somehow.

If your goal is to prove that a) is not always the case, then you should be looking for a counterexample. That is, find specific subspaces $Z,W$ of $\Bbb R^5$ where no element of $Z$ is orthogonal to $W$.

Your attempt for b):

ker(T)∩ W is not equal {0},,that mean kernel(T) =1 so by rank nullity theorem dim W of V =4 as we know that DimV = 5 and dim null(w) =1 so we get Dimrange(W) =4 so the option B is correct.

It is not true that kernel(T) = 1 since kernel(T) is a subspace an 1 is a number. It is also not necessarily true that $\dim \ker(T) = 1$.

Again, you're missing the fact that to correctly answer this question, you should construct an example. For instance: we can take $T:V \to V$ to be the orthogonal projection onto $V^\perp$.

For c): use the rank nullity theorem to deduce that $\dim \ker(T) \neq 0$. Now, if $T(x) = 0$, then we also have $S(T(x)) = 0$.