1. is Negative Definite Matrix ?
`[[-3,0,0],[0,-2,0],[0,0,-1]]`Solution:| `A` | = | | `-3` | `0` | `0` | | | `0` | `-2` | `0` | | | `0` | `0` | `-1` | |
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A matrix is negative definite if it's symmetric and all its pivots are negative.
Test method 1: Existence of all negative Pivots.First apply Gaussian Elimination method to find Pivots
| `A` | = | | `-3` | `0` | `0` | | | `0` | `-2` | `0` | | | `0` | `0` | `-1` | |
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Pivots are the first non-zero element in each row of this eliminated matrix.
`:.` Pivots are `-3,-2,-1`
Here all pivots are negative, so matrix is negative definite.
A matrix is negative definite if Determinants `D_i<0` for odd i and `D_i>0` for even i .
Test method 2: Determinants of all upper-left sub-matrices.| `A` | = | | `-3` | `0` | `0` | | | `0` | `-2` | `0` | | | `0` | `0` | `-1` | |
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| `-3` | `0` | `0` | | | `0` | `-2` | `0` | | | `0` | `0` | `-1` | |
| `=-6` |
Determinants are `-3,6,-6`
Here all odd determinants `D_1,D_3<0` and even determinants `D_2>0`, so matrix is negative definite.
A matrix is negative definite if it's symmetric and all its eigenvalues are negative.
Test method 3: All negative eigen values.`|A-lamdaI|=0`
| `(-3-lamda)` | `0` | `0` | | | `0` | `(-2-lamda)` | `0` | | | `0` | `0` | `(-1-lamda)` | |
| = 0 |
`:.(-3-lamda)((-2-lamda) × (-1-lamda) - 0 × 0)-0(0 × (-1-lamda) - 0 × 0)+0(0 × 0 - (-2-lamda) × 0)=0`
`:.(-3-lamda)((2+3lamda+lamda^2)-0)-0(0-0)+0(0-0)=0`
`:.(-3-lamda)(2+3lamda+lamda^2)-0(0)+0(0)=0`
`:. (-6-11lamda-6lamda^2-lamda^3)-0+0=0`
`:.(-lamda^3-6lamda^2-11lamda-6)=0`
`:.-(lamda+1)(lamda+2)(lamda+3)=0`
`:.(lamda+1)=0 or(lamda+2)=0 or(lamda+3)=0 `
`:.` The eigenvalues of the matrix `A` are given by `lamda=-3,-2,-1`,
Here all determinants are negative, so matrix is negative definite.
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