Question

Prove rank $I - uu^* = n-1$ (using ... row reduction?)

score 67 · Answer 1 · 2026-03-29 00:06:42

67

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Prove rank $I - uu^* = n-1$ (using ... row reduction?)

Published on 29 Mar 2026 - 0:06

score 54 · Answer 2 · 2022-03-05 08:50:21

54

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Finding the Rank of a special $n\times n$ matrix

Published on 05 Mar 2022 - 8:50

score 79 · Answer 3 · 2022-03-08 02:40:14

79

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Can sandwiching an invertible matrix change the rank or null space?

Published on 08 Mar 2022 - 2:40

score 158 · Answer 4 · 2022-03-11 17:19:52

158

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When 2 matrices commute and have the same rank?

Published on 11 Mar 2022 - 17:19

score 83 · Answer 5 · 2026-03-31 22:51:43

83

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if range $A^* \cap$ range $B^* = 0$ and $B^*A = 0$ then $\text{rank }(A + B) = \text{rank }A + \text{rank }B$

Published on 31 Mar 2026 - 22:51

score 64 · Answer 6 · 2022-03-15 11:48:07

64

Views

Is the rank of a matrix invariant upon removal of a column containing the same value?

Published on 15 Mar 2022 - 11:48

score 61 · Answer 7 · 2022-03-21 01:52:47

61

Views

A matrix rank problem

Published on 21 Mar 2022 - 1:52

score 192 · Answer 8 · 2026-02-23 12:34:37

192

Views

Explaining the non convexity of rank 1 matrix

Published on 23 Feb 2026 - 12:34

score 79 · Answer 9 · 2022-03-23 18:23:15

79

Views

I have a (wrong?) proof that a matrix with a unit eigenvalue can't have a zero eigenvalue. Where is the illegal step?

Published on 23 Mar 2022 - 18:23

score 72 · Answer 10 · 2022-03-25 09:59:25

72

Views

The matrix with entries $\pm 1$ is surjective?

Published on 25 Mar 2022 - 9:59

score 79 · Answer 11 · 2022-03-29 16:20:21

79

Views

Some questions of the proof related to Low-rank approximation on wikipedia

Published on 29 Mar 2022 - 16:20

score 77 · Answer 12 · 2022-03-31 13:25:19

77

Views

Is the rank $A+\sqrt{2} B$ greater than the rank of $A$?

Published on 31 Mar 2022 - 13:25

score 192 · Answer 13 · 2022-04-02 06:25:19

192

Views

Rank of a complex matrix

Published on 02 Apr 2022 - 6:25

score 627 · Answer 14 · 2022-04-03 20:16:17

627

Views

For any $A\in M_n(\mathbb{R})$, there is a $B$ s.t. $\operatorname{rank}(A)+\operatorname{rank}(B)=n$ and $AB=0$.

Published on 03 Apr 2022 - 20:16

score 133 · Answer 15 · 2026-03-28 03:26:23

133

Views

rank of a matrix and its transpose

Published on 28 Mar 2026 - 3:26

List Question

Prove rank $I - uu^* = n-1$ (using ... row reduction?)

Finding the Rank of a special $n\times n$ matrix

Can sandwiching an invertible matrix change the rank or null space?

When 2 matrices commute and have the same rank?

if range $A^* \cap$ range $B^* = 0$ and $B^*A = 0$ then $\text{rank }(A + B) = \text{rank }A + \text{rank }B$

Is the rank of a matrix invariant upon removal of a column containing the same value?

A matrix rank problem

Explaining the non convexity of rank 1 matrix

I have a (wrong?) proof that a matrix with a unit eigenvalue can't have a zero eigenvalue. Where is the illegal step?

The matrix with entries $\pm 1$ is surjective?

Some questions of the proof related to Low-rank approximation on wikipedia

Is the rank $A+\sqrt{2} B$ greater than the rank of $A$?

Rank of a complex matrix

For any $A\in M_n(\mathbb{R})$, there is a $B$ s.t. $\operatorname{rank}(A)+\operatorname{rank}(B)=n$ and $AB=0$.

rank of a matrix and its transpose

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