# determinant of rank-one update of identity matrix

I read something that suggests that if $I$ is the $n$-by-$n$ identity matrix, $v$ is an $n$-dimensional real column vector with $\|v\| = 1$ (standard Euclidean norm), and $t > 0$, then $\det(I+tvv^T)=1+t$. Can anyone prove this or provide a reference?

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$tvv^{T}$ is a number, not a matrix, so $I+tvv^{T}$ is undefined, let alone its determinant –  user123123 Oct 23 '12 at 23:41
No, $v$ is a column vector, so $vv^T$ is a square matrix. –  Stefan Smith Oct 23 '12 at 23:42
@Peter You're thinking of $\mathbf{v}^\mathrm{T}\mathbf{v}$. –  EuYu Oct 23 '12 at 23:43
Oops, sorry. Disregard what I said! –  user123123 Oct 23 '12 at 23:43
@bogus Congratulations. You should probably post that as an answer. –  EuYu Oct 23 '12 at 23:46

Rank one update, reference Matrix Analysis and Aplied Linear Algebra, Carl D. Meyer, page 475:

If $A_{n \times n}$ is nonsingular, and if $\mathbf{c}$ and $\mathbf{d}$ are $n \times 1$ columns, then $$\det(\mathbf{I} + \mathbf{c}\mathbf{d}^T) = 1 + \mathbf{d}^T\mathbf{c} \tag{6.2.2}$$ $$\det(A + \mathbf{c}\mathbf{d}^T) = \det(A)(1 + \mathbf{d}^T A^{-1}\mathbf{c}) \tag{6.2.3}$$

So in your case, $A=\mathbf{I}$ and the determinant is $1(1+ t\mathbf{v}^T\mathbf{v})=1+t$

EDIT. Further from the text:

Proof. The proof of (6.2.2) [the previous] follows by applying the product rules (p. 467) to $$\pmatrix{\mathbf{I} & \mathbf{0} \\ \mathbf{d}^T & 1}\pmatrix{\mathbf{I} + \mathbf{c}\mathbf{d}^T& \mathbf{c} \\ \mathbf{0} & 1}\pmatrix{\mathbf{I} & \mathbf{0} \\ -\mathbf{d}^T & 1}=\pmatrix{\mathbf{I} & \mathbf{c} \\ \mathbf{0} & 1 + \mathbf{d}^T\mathbf{c}}$$

To prove (6.2.3) write $A + \mathbf{c}\mathbf{d}^T = A ( \mathbf{I} + A^{-1}\mathbf{c}\mathbf{d}^T)$, and apply the product rule (6.1.15) along with (6.2.2)

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@navigetor23: for $\det(1+tvv^T)$ (if that is what you mean) take $\mathbf c=tc$ and $\mathbf d=v$. As for $A$, it is just an extra, to show one still has something similar if the identity is replaced by another invertible matrix. –  Marc van Leeuwen Oct 24 '12 at 10:11
@navigetor23 : I am the OP. adam W's answer contains a small mistake, but I accepted it, because, unlike mine, it shows what happens for a nonsymmetric rank-one update of the identity matrix. The reference is nice too. Thanks for the upvote. –  Stefan Smith Oct 24 '12 at 11:25

I solved it. The determinant of $I+tvv^T$ is the product of its eigenvalues. $v$ is an eigenvector with eigenvalue $1+t$. $I+tvv^T$ is real and symmetric, so it has a basis of real mutually orthogonal eigenvectors, one of which is $v$. If $w$ is another one, then $(I+tvv^T)w=w$, so all the other eigenvalues are $1$.

I feel like I should have known this already. Can anyone provide a reference for this and similar facts?

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Sylvester's determinant theorem states that more generally $$\det(I_k+AB)=\det(I_l+BA)$$ for any $k\times l$ matrix $A$ and $l\times k$ matrix $B$. You can apply this for $(k,l)=(n,1)$, $A=tv$ and $B=v^T$. See the link provided for a straightforward proof, using row and column operations.

In fact the proof is almost a one-liner, so here it is: in $(k+l)\times(k+l)$ block matrices one has $$\det\begin{pmatrix}I_k+AB&A\\0&I_l\end{pmatrix} =\det\begin{pmatrix}I_k&A\\-B&I_l \end{pmatrix} =\det\begin{pmatrix}I_k&A\\0&I_l+BA \end{pmatrix},$$ where the first equality is a compound column operation (subtract $B$ times the second block-column from the first, multiplication being on the right for column operations), and the second is a compound row operation (add $B$ times the first block-row to the second, multiplication being on the left for row operations), and the desired identity follows from computation of block-triangular determinants.

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Here's another proof (cf. Sherman–Morrison formula):

The non-zero eigenvalues of $AB$ and $BA$ are the same. This is straightforward to prove.

Hence the non-zero eigenvalues of $ab^T$ and $b^Ta$ are the same (that is, exactly one non-zero eigenvalue).

Hence the eigenvalues of $I+ ab^T$ are $1+b^Ta, 1,...,1$, and since the determinant is the product of eigenvalues, we have $\det(I+ab^T) = 1+b^Ta$.

In this particular example, $a=tv$, $b=v$, and $\|v\| = 1$, hence $b^Ta = t$, and so $\det(I+t v v^T) = 1+t$.

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