Mixing
Compare ratio of two determinants before and after adding sum of rank 1 matrices in both determinants
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I have a ratio between two determinants:
$$r_1 = frac{det(lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T)}{det(lambda I_d)}$$
where $x in mathbb{R}^{d}$, $||x||_{2} leq 1$, $lambda > 0$, and $I_d in mathbb{R}^{d times d}$ is identity matrix.
Now I add another sum of $n_p$ rank-1 matrices $sum_{t=1}^{n_p} x_{t}{x_{t}}^T$ (or equivalently write it as ${X_{p}}^{T}X_{p}$, where each row of $X_{p} in mathbb{R}^{n_{p} times d}$ is $x_t$) to both determinants.
$$r_2 = frac{det(lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T + {X_{p}}^{T}X_{p})}{det(lambda I_d + {X_{p}}^{T}X_{p})}$$
Now I want to compare the two ratios $r_1$ and $r_2$. Or maybe even better, with some other assumptions I can say $r_1$ is how many times of $r_2$.
Following the generalization of this matrix-determinant lemma, I can rewrite $r_2$ as:
$$r_2 = frac{det(A + {X_{p}}^{T}X_{p})}{det(lambda I_d + {X_{p}}^{T}X_{p})} = frac{det(A)det(I_{n_p}+X_{p} A^{-1} {X_{p}}^{T})}{det(lambda I_d)det(I_{n_p}+ X_{p} (lambda I_d)^{-1}{X_{p}}^{T})} = r_{1}frac{det(I_{n_p}+X_{p} A^{-1} {X_{p}}^{T})}{det(I_{n_p}+ X_{p} (lambda I_d)^{-1}{X_{p}}^{T})}$$
where $A=lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T$.
So the problem of comparing the two ratios becomes comparing $frac{det(I_{n_p}+X_{p} (lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T)^{-1} {X_{p}}^{T})}{det(I_{n_p}+ X_{p} (lambda I_d)^{-1}{X_{p}}^{T})}$ with $1$.
I don't know what to do from here. Intuitively the numerator should be smaller than denominator, since the term inside the "inverse" in the numerator is "bigger", so $r_2$ should be smaller.
linear-algebra matrices determinant positive-definite symmetric-matrices
asked Jan 31 at 19:59
Chuanhao LiChuanhao Li
132
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add a comment |
$begingroup$
I have a ratio between two determinants:
$$r_1 = frac{det(lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T)}{det(lambda I_d)}$$
where $x in mathbb{R}^{d}$, $||x||_{2} leq 1$, $lambda > 0$, and $I_d in mathbb{R}^{d times d}$ is identity matrix.
Now I add another sum of $n_p$ rank-1 matrices $sum_{t=1}^{n_p} x_{t}{x_{t}}^T$ (or equivalently write it as ${X_{p}}^{T}X_{p}$, where each row of $X_{p} in mathbb{R}^{n_{p} times d}$ is $x_t$) to both determinants.
$$r_2 = frac{det(lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T + {X_{p}}^{T}X_{p})}{det(lambda I_d + {X_{p}}^{T}X_{p})}$$
Now I want to compare the two ratios $r_1$ and $r_2$. Or maybe even better, with some other assumptions I can say $r_1$ is how many times of $r_2$.
Following the generalization of this matrix-determinant lemma, I can rewrite $r_2$ as:
$$r_2 = frac{det(A + {X_{p}}^{T}X_{p})}{det(lambda I_d + {X_{p}}^{T}X_{p})} = frac{det(A)det(I_{n_p}+X_{p} A^{-1} {X_{p}}^{T})}{det(lambda I_d)det(I_{n_p}+ X_{p} (lambda I_d)^{-1}{X_{p}}^{T})} = r_{1}frac{det(I_{n_p}+X_{p} A^{-1} {X_{p}}^{T})}{det(I_{n_p}+ X_{p} (lambda I_d)^{-1}{X_{p}}^{T})}$$
where $A=lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T$.
So the problem of comparing the two ratios becomes comparing $frac{det(I_{n_p}+X_{p} (lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T)^{-1} {X_{p}}^{T})}{det(I_{n_p}+ X_{p} (lambda I_d)^{-1}{X_{p}}^{T})}$ with $1$.
I don't know what to do from here. Intuitively the numerator should be smaller than denominator, since the term inside the "inverse" in the numerator is "bigger", so $r_2$ should be smaller.
linear-algebra matrices determinant positive-definite symmetric-matrices
asked Jan 31 at 19:59
Chuanhao LiChuanhao Li
132
$endgroup$
add a comment |
$begingroup$
I have a ratio between two determinants:
$$r_1 = frac{det(lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T)}{det(lambda I_d)}$$
where $x in mathbb{R}^{d}$, $||x||_{2} leq 1$, $lambda > 0$, and $I_d in mathbb{R}^{d times d}$ is identity matrix.
Now I add another sum of $n_p$ rank-1 matrices $sum_{t=1}^{n_p} x_{t}{x_{t}}^T$ (or equivalently write it as ${X_{p}}^{T}X_{p}$, where each row of $X_{p} in mathbb{R}^{n_{p} times d}$ is $x_t$) to both determinants.
$$r_2 = frac{det(lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T + {X_{p}}^{T}X_{p})}{det(lambda I_d + {X_{p}}^{T}X_{p})}$$
Now I want to compare the two ratios $r_1$ and $r_2$. Or maybe even better, with some other assumptions I can say $r_1$ is how many times of $r_2$.
Following the generalization of this matrix-determinant lemma, I can rewrite $r_2$ as:
$$r_2 = frac{det(A + {X_{p}}^{T}X_{p})}{det(lambda I_d + {X_{p}}^{T}X_{p})} = frac{det(A)det(I_{n_p}+X_{p} A^{-1} {X_{p}}^{T})}{det(lambda I_d)det(I_{n_p}+ X_{p} (lambda I_d)^{-1}{X_{p}}^{T})} = r_{1}frac{det(I_{n_p}+X_{p} A^{-1} {X_{p}}^{T})}{det(I_{n_p}+ X_{p} (lambda I_d)^{-1}{X_{p}}^{T})}$$
where $A=lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T$.
So the problem of comparing the two ratios becomes comparing $frac{det(I_{n_p}+X_{p} (lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T)^{-1} {X_{p}}^{T})}{det(I_{n_p}+ X_{p} (lambda I_d)^{-1}{X_{p}}^{T})}$ with $1$.
I don't know what to do from here. Intuitively the numerator should be smaller than denominator, since the term inside the "inverse" in the numerator is "bigger", so $r_2$ should be smaller.
linear-algebra matrices determinant positive-definite symmetric-matrices
asked Jan 31 at 19:59
Chuanhao LiChuanhao Li
132
$endgroup$
I have a ratio between two determinants:
$$r_1 = frac{det(lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T)}{det(lambda I_d)}$$
where $x in mathbb{R}^{d}$, $||x||_{2} leq 1$, $lambda > 0$, and $I_d in mathbb{R}^{d times d}$ is identity matrix.
Now I add another sum of $n_p$ rank-1 matrices $sum_{t=1}^{n_p} x_{t}{x_{t}}^T$ (or equivalently write it as ${X_{p}}^{T}X_{p}$, where each row of $X_{p} in mathbb{R}^{n_{p} times d}$ is $x_t$) to both determinants.
$$r_2 = frac{det(lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T + {X_{p}}^{T}X_{p})}{det(lambda I_d + {X_{p}}^{T}X_{p})}$$
Now I want to compare the two ratios $r_1$ and $r_2$. Or maybe even better, with some other assumptions I can say $r_1$ is how many times of $r_2$.
Following the generalization of this matrix-determinant lemma, I can rewrite $r_2$ as:
$$r_2 = frac{det(A + {X_{p}}^{T}X_{p})}{det(lambda I_d + {X_{p}}^{T}X_{p})} = frac{det(A)det(I_{n_p}+X_{p} A^{-1} {X_{p}}^{T})}{det(lambda I_d)det(I_{n_p}+ X_{p} (lambda I_d)^{-1}{X_{p}}^{T})} = r_{1}frac{det(I_{n_p}+X_{p} A^{-1} {X_{p}}^{T})}{det(I_{n_p}+ X_{p} (lambda I_d)^{-1}{X_{p}}^{T})}$$
where $A=lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T$.
So the problem of comparing the two ratios becomes comparing $frac{det(I_{n_p}+X_{p} (lambda I_d + sum_{t=1}^{n}x_{t}{x_{t}}^T)^{-1} {X_{p}}^{T})}{det(I_{n_p}+ X_{p} (lambda I_d)^{-1}{X_{p}}^{T})}$ with $1$.
I don't know what to do from here. Intuitively the numerator should be smaller than denominator, since the term inside the "inverse" in the numerator is "bigger", so $r_2$ should be smaller.
linear-algebra matrices determinant positive-definite symmetric-matrices
linear-algebra matrices determinant positive-definite symmetric-matrices
asked Jan 31 at 19:59
Chuanhao LiChuanhao Li
132
asked Jan 31 at 19:59
Chuanhao LiChuanhao Li
132
asked Jan 31 at 19:59
Chuanhao LiChuanhao Li
132
asked Jan 31 at 19:59
Chuanhao LiChuanhao Li
132
asked Jan 31 at 19:59
Chuanhao LiChuanhao Li
132
132
add a comment |
add a comment |
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