MHB 4x4 Matrix with rank B=4 and B^2=3

[Petrus]

Hello MHB,
"Can we construct a $$4x4$$ Matrix $$B$$ so that rank $$B=4$$ but rank $$B^2=3$$"
My thought:
we got one condition for this to work is that det $$B=0$$ and det $$B^2 \neq 0$$ and B also have to be a upper/lower or identity Matrix. And this Will not work.. I am wrong or can I explain this in a better way?

Regards,
$$|\pi\rangle$$

 

[Evgeny.Makarov]

"Can we construct a $$4x4$$ Matrix $$B$$ so that rank $$B=4$$ but rank $$B^2=3$$"
My thought:
we got one condition for this to work is that det $$B=0$$ and det $$B^2 \neq 0$$

It's the other way around: $\mathop{\text{rank}}(B)=4\iff\det(B)\ne0$ and $\mathop{\text{rank}}(B^2)=3\implies\det(B^2)=0$. But you are right that this is impossible because $\det(B^2)=(\det(B))^2$.

 

[Petrus]

It's the other way around: $\mathop{\text{rank}}(B)=4\iff\det(B)\ne0$ and $\mathop{\text{rank}}(B^2)=3\implies\det(B^2)=0$. But you are right that this is impossible because $\det(B^2)=(\det(B))^2$.

Hello Evgeny.Makarov,
thanks for fast respond and I meant that! And thanks for showing me this I did not know that $\det(B^2)=(\det(B))^2$ That Was what I Was looking for

Regards,
$$|\pi\rangle$$

 

[Fernando Revilla]

An alternative proof (without using determinants):

Consider the linear map $B:\mathbb{R}^4\to \mathbb{R}^4,\; x\to Bx.$ As $\operatorname{rank}B=4,$ $\operatorname{nullity}B=0,$ which implies $B$ is bijective. But the composition of bijective maps is bijective, so $\operatorname{rank}B^2=4.$

 

[Deveno]

Another formulation:

As rank(B) = 4, B is surjective, that is, B(R⁴) = R⁴ (for this is what rank means: the dimension of the image space (or column space) of B, and R⁴ is the ONLY 4-dimensional subspace of R⁴​).

Consequently, B²(R⁴) = B(B(R⁴)) = B(R⁴) = R⁴, from which we conclude B² is likewise surjective, and thus rank(B²) = 4 as well.

(I only post this to indicate one need not even invoke the rank-nullity theorem).

 

[Petrus]

this is impossible because $\det(B^2)=(\det(B))^2$.

Now that I think about I remember a sats that said $$|AB|=|A||B|$$ but in this case $$A=B$$ hmm I need to find the proof for this.

Regards,
$$|\pi\rangle$$

 

[Deveno]

You can find various proofs and references here:

homework - How to show $\det(AB) =\det(A)\det(B)$ - Mathematics Stack Exchange

As I recall, any proof of this is somewhat long-winded and rather tedious. Determinants can get rather complicated as the size of matrices goes up (because of the permutative nature of them).

 

[Evgeny.Makarov]

I saw one neat proof of $\det(AB)=\det(A)\det(B)$ in Linear Algebra and Its Applications by Gilbert Strang. He defines $\det(\cdot)$ as a function satisfying three properties:

(1) $\det(I)=1$ where $I$ is the identity matrix;
(2) it changes sign when two adjacent rows are swapped;
(3) it is linear on the first row.

Signed volume in an orthonormal basis satisfies these properties, so this definition is much more intuitive than the Leibniz formula, which is derivable from (1)–(3).

Now, to prove that $\det(AB)=\det(A)\det(B)$, fix $B$ and consider $d(A)=\det(AB)/\det(B)$. It is possible to show that $d(A)$ satisfies (1)–(3), and so $d(A)=\det(A)$.

Now that I looked at the StackExchange link, this is answer #2, which is highest-ranked.