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Home > Matrix & Vector calculators > LQ Decomposition example
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15. LQ Decomposition example
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- Example `[[1,-1,4],[1,4,-2],[1,4,2],[1,-1,0]]`
- Example `[[3,-6],[4,-8],[0,1]]`
- Example `[[1,-4],[2,3],[2,2]]`
- Example `[[1,2,4],[0,0,5],[0,3,6]]`
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Other related methods
- Transforming matrix to Row Echelon Form (ref)
- Transforming matrix to Reduced Row Echelon Form (rref)
- Rank of matrix
- Characteristic polynomial of matrix
- Eigenvalues
- Eigenvectors (Eigenspace)
- Triangular Matrix
- LU decomposition using Gauss Elimination method of matrix
- LU decomposition using Doolittle's method of matrix
- LU decomposition using Crout's method of matrix
- Diagonal Matrix
- Cholesky Decomposition
- QR Decomposition (Gram Schmidt Method)
- QR Decomposition (Householder Method)
- LQ Decomposition
- Pivots
- Singular Value Decomposition (SVD)
- Moore-Penrose Pseudoinverse
- Power Method for dominant eigenvalue
- Inverse Power Method for dominant eigenvalue
- Determinant by gaussian elimination
- Expanding determinant along row / column
- Determinants using montante (bareiss algorithm)
- Leibniz formula for determinant
- determinants using Sarrus Rule
- determinants using properties of determinants
- Row Space
- Column Space
- Null Space
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2. Example `[[3,-6],[4,-8],[0,1]]`
(Previous example) | 4. Example `[[1,2,4],[0,0,5],[0,3,6]]`
(Next example) |
3. Example `[[1,-4],[2,3],[2,2]]`
Find LQ Decomposition ...
`[[1,-4],[2,3],[2,2]]`Solution:Suppose you want a `LQ` factorization of a matrix `A`, then you do a QR factorization of `A^T`, i.e., `A^T=UR`, where `U` is orthogonal and `R` is upper triangular. Then `A=LQ=R^TU^T` where `L=R^T` is lower triangular, and `Q=U^T` is orthogonal. Now, UR Decomposition of `A'` by GramSchmidt Method `r_(11)=||q_1'||=sqrt((1)^2+(-4)^2)=sqrt(17)=4.1231056256` | `q_1 = 1/(||q_1'||) * q_1'` | = | `1/4.1231056256 * ` | | = | | `0.242535625` | | | `-0.9701425001` | |
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| `r_(12)=q_1^T * a_2` | = | | [ | `0.242535625` | `-0.9701425001` | ] |
| `xx` | | `=-2.4253562504` |
| `q_2'` | `=a_2-r_(12) * q_1` | = | | `+2.4253562504` | | `0.242535625` | | | `-0.9701425001` | |
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| = | | `2.5882352941` | | | `0.6470588235` | |
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`r_(22)=||q_2'||=sqrt((2.5882352941)^2+(0.6470588235)^2)=sqrt(7.1176470588)=2.6678918754` | `q_2 = 1/(||q_2'||) * q_2'` | = | `1/2.6678918754 * ` | | `2.5882352941` | | | `0.6470588235` | |
| = | | `0.9701425001` | | | `0.242535625` | |
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| `r_(13)=q_1^T * a_3` | = | | [ | `0.242535625` | `-0.9701425001` | ] |
| `xx` | | `=-1.4552137502` |
| `r_(23)=q_2^T * a_3` | = | | [ | `0.9701425001` | `0.242535625` | ] |
| `xx` | | `=2.4253562504` |
| `q_3'` | `=a_3-r_(13) * q_1-r_(23) * q_2` | = | | `+1.4552137502` | | `0.242535625` | | | `-0.9701425001` | |
| `-2.4253562504` | | `0.9701425001` | | | `0.242535625` | |
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| = | |
`r_(33)=||q_3'||=sqrt((0)^2+(0)^2)=sqrt(0)=0` | `q_3 = 1/(||q_3'||) * q_3'` | = | `1/0 * ` | | = | |
| `U` | `=[q_1,q_2,q_3]` | = | | `0.242535625` | `0.9701425001` | `0.8` | | | `-0.9701425001` | `0.242535625` | `0.6` | |
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| `R` | = | | `r_(11)` | `r_(12)` | `r_(13)` | | | `0` | `r_(22)` | `r_(23)` | | | `0` | `0` | `r_(33)` | |
| = | | `4.1231056256` | `-2.4253562504` | `-1.4552137502` | | | `0` | `2.6678918754` | `2.4253562504` | | | `0` | `0` | `0` | |
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Now, L and Q from R and U | `L=R^T` | = | | `4.1231056256` | `0` | `0` | | | `-2.4253562504` | `2.6678918754` | `0` | | | `-1.4552137502` | `2.4253562504` | `0` | |
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| `Q=U^T` | = | | `0.242535625` | `-0.9701425001` | | | `0.9701425001` | `0.242535625` | | | `0.8` | `0.6` | |
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checking `L xx Q = A?` | `L xx Q` | = | | `4.1231056256` | `0` | `0` | | | `-2.4253562504` | `2.6678918754` | `0` | | | `-1.4552137502` | `2.4253562504` | `0` | |
| `xx` | | `0.242535625` | `-0.9701425001` | | | `0.9701425001` | `0.242535625` | | | `0.8` | `0.6` | |
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Solution is possible.
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
2. Example `[[3,-6],[4,-8],[0,1]]`
(Previous example) | 4. Example `[[1,2,4],[0,0,5],[0,3,6]]`
(Next example) |
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