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This problem comes from exercise from my lecturer

Let $A$ be a square matrix such that $\mbox{rank}(A)=1$. Prove that $\det(A+I)=\mbox{trace}(A)+1$.

($I$ is the identy matrix)

I usually haven't any idea in solving linear algebra problem.

Julien
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Yoshua Yonatan
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4 Answers4

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Edited in response to julien's comment

Recall that:

1) the rank is greater than or equal to the number of non-zero eigenvalues

2) the determinant is equal to the product of all eigenvalues

3) the trace is equal to the sum of all eigenvalues

You should also be able to show that if $\lambda$ is an eigenvalue of $A$ then $\lambda+1$ is an eigenvalue of $A+I$

I think you should be able to do the proof now. Let me know if you need more help.

john
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    What about $A=\pmatrix{0&1\ 0&0}$? Your argument works in the diagonalizable case, which is equivalent to the trace being nonzero for rank one square matrices. – Julien Jun 25 '13 at 13:25
  • Oops! Thanks for pointing that out Julien. I believe that should be fixed now and we can just separate into two cases- one non-zero eigenvalue or no non-zero eigenvalues, both of which are then simple to complete the proof for. – john Jun 25 '13 at 13:42
  • Yes. You are implicitly using Jordan normal form and assuming that the base field $K$ be algebraically closed for these 3 statements to make sense in full generality. But in the rank one case, the characteristic polynomial splits over $K$ whether it is algebraically closed or not, so that's ok. But note that the crux of this approach is precisely the fact that the characteristic polynomial splits. – Julien Jun 25 '13 at 13:46
  • @john Could you please write the solution as I am having difficulty getting the conclusion? – user444042 Nov 23 '17 at 15:58
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Hint: If the rank of the matrix is $1$, it means that all columns can be written as multiples of one of them (say, wlog, the first one); since the determinant is invariant by linear combination of the columns, you can write $\det(A+I)$ as $\det(A'+I)$, where $A'$ and $A$ have same rank, but only the first column of $A'$ is non-zero.

Clement C.
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  • Perhaps there is an easy way to see Clement's answer by column operations, but if you think of it in terms of bases I would add that the problem hinges on the fact that $I$ has the same matrix with respect to any basis. – Eric Auld Jun 25 '13 at 13:23
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In a suitable base, $A$ is a matrix $a e_{ij}$, where $a \ne 0$, and as usual $e_{ij}$ is the matrix that has all entries zero, except the $(i,j)$ one, which is $1$.

(Whereas, as noted by @EricAuld, the identity map has matrix $I$ with respect to any base.)

There are two cases to distinguish.

  1. When $i \ne j$, then $1 + \operatorname{trace}(A) = 1 = \det(I + A)$.
  2. When $i = j$, then $1 + \operatorname{trace}(A) = 1 + a = \det(I + A)$.

Proof of the first statement. The image of $A$ has dimension $1$. Choose a non-zero vector $v$ in the image.

Now if $A v \ne 0$, complete $v$ to a base of the whole space by adding a base of $\ker(A)$, which has dimension $n-1$, if the matrix is $n \times n$. This is case (2).

If $A v = 0$, choose an element $u$ such that $A u \ne 0$, and take as a base $u, v$ and $n-2$ further vectors that together with $v$ form a base of $\ker(A)$. This is case (1).

Andreas Caranti
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An example.

Let $A$ be the $2\times 2$ matrix $A=(A_1,A_2)$, with $A_1=A_2=(1,1)^t$. So $tr(A)=2$ and $\det(A+I)=3=tr(A)+1$.

EDIT: the original formula for the determinant contained a mistake .Should I remove the answer? Without a more general proof it does not support the OP. Thanks

Avitus
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