# Which of these linear algebra statements are true?

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..

• a) If W=Z this is clearly unachieveable Commented Aug 24, 2017 at 12:14
• how can u say that it is unachievable @ aleksejs Fomin
– user469754
Commented Aug 24, 2017 at 12:17
• You should post using some basic MathJax. Commented Aug 24, 2017 at 12:18
• If a vector is orthogonal to the entire space, it is orthogonal to all vectors in it. If W=Z, the vector z will be at least certainly not orthogonal to itself in W Commented Aug 24, 2017 at 12:19
• Commented Aug 24, 2017 at 12:21

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 $\ker(S\circ T)\neq\{0\}$.

• im not getting rank T= 1 for opton b)....and for option c)there exists x ∈ V such that x is not equal 0 ....and .how can u write ker( S ◦ T) = 0,,,im not getting this line @ jose carlos santos
– user469754
Commented Aug 24, 2017 at 12:37
• @legohlegoh What is it that you don't get for option b)? Concerning c), $\det(S\circ T)=0\iff\operatorname{rank}(S\circ T)<\dim V$. Commented Aug 24, 2017 at 12:39
• for option b) kernel T= 1 that mean Rank T = 4 by rank nullity theorem @ jose carlo santos
– user469754
Commented Aug 24, 2017 at 12:43
• @legohlegoh Perhaps that you are not aware of the fact that $\operatorname{rank}T=\dim\operatorname{im}T$. Commented Aug 24, 2017 at 12:49
• every subspace W of V of dimension 4. ...here dimension of the image of T =4 ,,,then how can u write dim imT =1 @jose carlos santos
– user469754
Commented Aug 24, 2017 at 13:00

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 $W$.

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 $S(T(x)) = 0$.

• thanks a lot @Omnomnomnom for rectifying my mistakes...
– user469754
Commented Aug 24, 2017 at 12:40

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

• As subspaces always contain the zero vector, your argument for b) will not work.
– Dirk
Commented Aug 24, 2017 at 12:26
• Ah I guess I misunderstood the question, sorry :) I thought that they did not intersect at all Commented Aug 24, 2017 at 12:27
• You can't have two parallel spaces that both contain the zero vector, as parallel spaces shouldn't intersect. Try drawing two parallel lines in $\mathbb{R}^2$ that both contain the zero point but are not equal.
– Dirk
Commented Aug 24, 2017 at 12:29