This is part of our concern have positive eigenvalues of symmetric matrix. If the symmetry of a matrix so that important, then all eigenvalues greater than zero this additional property to let this matrix is truly exceptional. But here is not the particular rare, in fact symmetric matrix with positive eigenvalues is very common in various applications, they are called positive definite matrix .
We can by checking the value is greater than zero feature to identify positive definite matrix, but the calculated value is a characteristic work, when we really need them, we can calculate, but if we just want to know if they are positive, we have faster The way.
1. Analyzing definite matrices
First, since the matrix is symmetric, all the eigenvalues are real numbers NATURAL. Let's start with a matrix of 2 × 2,
\[A = \begin{bmatrix} a&b \\b&c\end{bmatrix}\]
A characteristic value is positive if and only if \ (a> 0 \) and \ (AC-2 ^ B> 0 \) .
If the characteristic value of 2 × 2 matrix \ (\ lambda_1> 0 \) , \ (\ lambda_2> 0 \) , then the determinant is equal to their product, \ (\ lambda_1 \ lambda_2 = | A | = 2 ^ B-AC > 0 \) , which is equal to the matrix trace, \ (\ lambda_1 + \ lambda_2 = a + C> 0 \) , so \ (a \) and \ (C \) must be positive.
A characteristic value is positive if and only if the primary is positive.
This connection of the two parts of linear algebra, a positive value means that the feature of main positive element, and vice versa . Further, the main element is often faster than the characteristic value calculation.
- Based on the definition of energy
\[Ax=\lambda x \to x^TAx=\lambda x^Tx=\lambda ||x||^2>0\]
Therefore, if the feature value is greater than zero, \ (X ^ TAx \) for all feature vectors is also greater than zero. In fact, not just a feature vector, for any nonzero vector \ (the X-\) , the above equation is also true.
A is positive definite, if there \ (x ^ TAx> 0 \ ) for any nonzero vector are true.
From this definition, we can conclude that if \ (A, B \) is a symmetric positive definite matrix, then \ (A + B \) also.
If \ (R & lt \) columns are uncorrelated, then the \ (A = R ^ TR \ ) is positive definite.
\ [X ^ = x ^ TR South TRX = ^ (Rx) = || Rx TRX ^ || ^ 2 \]
Since \ (R & lt \) column is irrelevant, so that for any nonzero vector \ (X \) , \ (the Rx \ = Not \ {0} boldsymbol \) .
When a symmetrical matrix having one of the following five attributes, then it must satisfy all of the properties.
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- All \ (n-\) th primary is positive.
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- All \ (n-\) th upper left determinant is positive, that is, \ (1 × 1, 2 × 2 \ cdots n × n \) determinant.
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- All \ (n-\) eigenvalues are positive.
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- \ (x ^ TAx> 0 \ ) except a zero vector.
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- \ (A = R ^ TR \ ) to a column matrix has uncorrelated \ (R & lt \) .
2. semi-definite matrix
We will often definite edge, the determinant is zero, the smallest eigenvalue is zero, the edges of these matrix called semi positive definite matrices.
\ (A \) of the characteristic value is 0 and 5, 1 and 0 for the upper left determinant, which has rank 1, can be decomposed into a matrix with associated columns \ (R & lt TR ^ \) .
If you add a 4 element arbitrarily small number, it will become positive definite matrix. Likewise, \ (B \) may be written as \ (R ^ TR \) form, but \ (R & lt \) columns must be related.
3. The first application: oval \ (ax ^ 2 + 2bxy + cy ^ 2 = 1 \)
- Elliptical and inclined linked matrix A, \ (= X ^ TAx. 1 \) .
- Good oval matrix row and \ (\ the Lambda \) together, \ (X-T ^ \ = X-the Lambda. 1 \) .
- The elliptical rotation matrix is properly arranged eigenvectors \ (Q \) .
For elliptic equations \ (5X + 2 ^ 2 = ^ 8xy 5Y. 1 + \) , we have:
\[ \begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} 5 &4 \\ 4& 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = 1 \quad A = \begin{bmatrix} 5 &4 \\ 4& 5 \end{bmatrix} \]
The \ (A \) is decomposed into \ (Q \ Lambda Q ^ T \) we get:
\[\begin{bmatrix} 5 &4 \\ 4& 5 \end{bmatrix} = \frac{1}{\sqrt 2}\begin{bmatrix} 1&1 \\ 1&-1 \end{bmatrix} \begin{bmatrix} \boldsymbol 9&0 \\ 0&\boldsymbol 1 \end{bmatrix} \frac{1}{\sqrt 2} \begin{bmatrix} 1&1 \\ 1&-1 \end{bmatrix} \]
The elliptic equation may be rewritten as:
\[5x^2+8xy+5y^2=1 = 9*(\frac{x+y}{\sqrt 2})^2+1*(\frac{x-y}{\sqrt 2})^2\]
可以看到，方程的系数是两个特征值 9 和 0，而在平方内部则是两个特征向量 \((1, 1)/\sqrt 2\) 和 \((1, -1)/\sqrt 2\)。椭圆的坐标轴是沿着特征向量的方向，这也就是为什么 \(A=Q\Lambda Q^T\) 被称作主轴定理，特征向量指出了坐标轴的方向，特征值则指出了长度。
将椭圆排好后，较大的特征值 9 给出了短半轴的长度 \(1/\sqrt \lambda_1 = 1/3\)，较小的特征值 1 给出了长半轴的长度 \(1/\sqrt \lambda_2 = 1\)。在 \(xy\) 系统中，坐标轴沿着 \(A\) 的特征向量的方向，而在 \(XY\) 系统中，坐标轴沿着 \(\Lambda\) 的特征向量的方向。
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