2

The (maximum singular value) matrix norm constraint $\|A(x)\|<1$, where $A(x) \in \mathbb R^{p \times q}$ depends affinely on $x$, is represented as the following linear matrix inequality (LMI)

$$\begin{bmatrix} I & A(x) \\ A(x)^T & I \\ \end{bmatrix} >0$$

since $\|A(x)\|<1$ is equivalent to $I-AA^T > 0$. Why?

I would like to understand this statement of LMI. Thank you for your help and time.

Rodrigo de Azevedo
  • 1
Rosa Maria Gtz.
  • 3,756
  • What's the other side of the inequality here? – Ben Grossmann Jul 18 '16 at 12:04

  • 0 I edites, sorry.

     – Rosa Maria Gtz. Jul 18 '16 at 12:23
  • What's a LMI? (Not all abbreviations are 100% standard.) – Ian Jul 18 '16 at 12:58
  • Linear matrix inequalities – Rosa Maria Gtz. Jul 18 '16 at 12:59
  • 1
    So in other words, $\begin{bmatrix} I & A \ A^T & I \end{bmatrix}$ is positive definite? Then I can tell you how to handle the second part: $| A |_2$ is the largest singular value, which is the square root of the largest eigenvalue of $A A^T$. If $| A |_2<1$ then all eigenvalues of $A A^T$ are less than $1$ so all eigenvalues of $I-AA^T$, which are $1$ minus eigenvalues of $A A^T$, are positive. – Ian Jul 18 '16 at 13:01
  • Bus can I see $I-AA^T$ as the determinant of that matrix positive definite? – Rosa Maria Gtz. Jul 18 '16 at 16:37
  • The determinant is a number, regardless of writing things as block matrices. A different way to view it is to think of this as $I$ plus the positive definite matrix $\begin{bmatrix} 0 & A^T \ A & 0 \end{bmatrix}$ whose eigenvalues are the singular values of $A$. – Ian Jul 18 '16 at 16:40
  • Ok, just a merely question, if A is a square matrix and symmetric the A is symmetric ? I'm sure that it is true but I would like know your opinion , ty. – Rosa Maria Gtz. Jul 18 '16 at 19:39
  • Related: https://math.stackexchange.com/q/1963711/339790 – Rodrigo de Azevedo May 11 '18 at 11:06

2 Answers2

1

because $\begin{bmatrix} I & A(x) \\ A(x)^T & I \\ \end{bmatrix} $ is positive definite. For any non-empty vector,

$$\begin{bmatrix} z_1 &z_2 \end{bmatrix} \begin{bmatrix} I & A(x) \\ A(x)^T & I \\ \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix}$$

$$= ||z_1||^2+||z_2||^2+2||z_1^TA^Tz_2||$$ $$= ||z_1||^2+||z_2||^2+2||z_1||\cdot||A^TA||\cdot||z_2|| \gt 0$$

user115350
  • 1,291
1

My previous answer is wrong. Here is the correct one.

$$\begin{bmatrix} I & A(x) \\ A(x)^T & I \\ \end{bmatrix}= \begin{bmatrix} I & 0 \\ A(x)^T & I \\ \end{bmatrix} \begin{bmatrix} I & 0 \\ 0 & I-A(x)^TA(x) \\ \end{bmatrix} \begin{bmatrix} I & A(X) \\ 0 & I \\ \end{bmatrix} $$

Thus $\begin{bmatrix} I & A(x) \\ A(x)^T & I \\ \end{bmatrix} \gt 0$ is equivalent to $\begin{bmatrix}I-A(x)^TA(x)\end{bmatrix} \gt 0$ and @Ian in his comment proved the second part.

user115350
  • 1,291