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Finding rank of this matrix : $$A_{n\times n}= \begin{bmatrix} a+1&a&\cdots&a\\ a&a+1&\cdots&a\\ \vdots&\vdots&\ddots&\vdots\\ a&a&\cdots&a+1 \end{bmatrix} $$ $(a\geq 2, a\in \mathbb{N})$ I'm trying to prove $det(A) \ne 0$ so that $rank(A)=n$ but I don't know how to prove it quickly. Can anyone help me ?

user307596
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  • @SassatelliGiulio sr, i forgot to write conditions of a – user307596 Nov 27 '23 at 11:26
  • The sum of each row is same, so 'the sum' could be took out. Then you will find the determinant by simple substraction by rows. – Cunyi Nan Nov 27 '23 at 11:27
  • @CunyiNan still not get it, can u explain it ? – user307596 Nov 27 '23 at 11:28
  • Note: An especially efficient method to determine the value of a determinant of order $n$ can be obtained if we transform it in the same way we do in order to determine the rank of a matrix, i.e., all the elements under the diagonal $a_{11}, a_{22}, \ldots a_{nn}$ are equal to zero. Then the value of the determinant is the product of the diagonal elements of the transformed determinant. – Anton Vrdoljak Nov 27 '23 at 11:33

1 Answers1

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Here is an alternative approach. One has $$A = I_n + B$$

where $I_n$ is the identity matrix, and $B$ is the matrix whose all coefficients are equal to $a$. Since all the rows of $B$ are equal and $a \neq 0$, then $B$ has rank $1$, so the eigenvalues of $B$ are $0$ (with multiplicity $n-1$) and $\mathrm{Tr}(B)=na$ (with multiplicity $1$).

So the eigenvalues of $A$ are $1$ (with multiplicity $n-1$) and $na+1$ (with multiplicity $1$). In particular, $$\det(A)=1^{n-1} \times (na+1)^1 = na+1, \quad \text{i.e.} \quad \boxed{\det(A)=na+1}$$

Since $\det(A) \neq 0$, you get that the rank of $A$ is equal to $n$.

TheSilverDoe
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