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

Petrus

Well-known member
Hello MHB,
"Can we construct a $$\displaystyle 4x4$$ Matrix $$\displaystyle B$$ so that rank $$\displaystyle B=4$$ but rank $$\displaystyle B^2=3$$"
My thought:
we got one condition for this to work is that det $$\displaystyle B=0$$ and det $$\displaystyle B^2 \neq 0$$ 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,
$$\displaystyle |\pi\rangle$$

Evgeny.Makarov

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

Well-known member
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,
$$\displaystyle |\pi\rangle$$

Last edited:

Fernando Revilla

Well-known member
MHB Math Helper
An alternative proof (without using determinants):

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

Deveno

Well-known member
MHB Math Scholar
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

Well-known member
this is impossible because $\det(B^2)=(\det(B))^2$.
Now that I think about I remember a sats that said $$\displaystyle |AB|=|A||B|$$ but in this case $$\displaystyle A=B$$ hmm I need to find the proof for this.

Regards,
$$\displaystyle |\pi\rangle$$

MHB Math Scholar

Evgeny.Makarov

Well-known member
MHB Math Scholar
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.