Intuition & use of M*M^T product of matrix & its transpose?

In summary, the conversation discusses the use and intuition of multiplying a matrix by its transpose. The product is useful in cases where inner products are involved and can be seen as a way to dot the column vectors of the matrix together. In some cases, the product results in an identity matrix, while in others, it depends on the context.
  • #1
NotASmurf
150
2
Hi all, I've occasionly seen people multiply a matrix by its transpose, what is the use and intuition of the product? Any help appreciated.
 
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  • #2
Were these orthogonal matrices, where M-1 = MT?
 
  • #3
In one case yes, What would that mean intuition wise?
 
  • #4
NotASmurf said:
In one case yes,
In that case, its use is obvious, as M MT = I. Otherwise, it depends on the context.
 
  • #5
DrClaude said:
In that case, its use is obvious, as M MT = I. Otherwise, it depends on the context.
Was in stats, with covarience matrices.
 
  • #6
NotASmurf said:
Hi all, I've occasionly seen people multiply a matrix by its transpose, what is the use and intuition of the product? Any help appreciated.
These products show up when inner products are involved. For example, if you write elements of ##\mathbb R^n## as n×1 matrices, you can write ##x\cdot y=x^Ty##. From this you get results like ##(Mx)\cdot(My)=(Mx)^T(My)=x^TM^TMy##.
 
  • #7
What would that mean intuition wise?

The column vectors are orthonormal. Multiplying by the transpose makes you dot all the column vectors together with each other to get each entry of the product, which it the identity matrix.
 

1. What is the intuition behind using the M*M^T product of a matrix and its transpose?

The M*M^T product of a matrix and its transpose is used to obtain the square of the norm of the matrix. This is because it essentially multiplies each row of the matrix by its own transpose, resulting in a symmetric matrix. The diagonal elements of this symmetric matrix represent the squared norms of the original matrix's rows, while the off-diagonal elements represent the inner products between different rows. Thus, taking the trace of this matrix (the sum of the diagonal elements) gives the sum of the squared norms of the rows, i.e. the square of the norm of the original matrix.

2. How is the M*M^T product related to the concept of orthogonality?

The M*M^T product is related to orthogonality because it is used to check if a matrix is orthogonal, i.e. if its columns (or rows) are mutually orthogonal unit vectors. If a matrix A is orthogonal, then A*A^T will result in an identity matrix, since the inner product between any two different columns (or rows) will be 0. Thus, the M*M^T product can be used to verify the orthogonality of a matrix.

3. Can the M*M^T product of a matrix and its transpose be used for dimensionality reduction?

Yes, the M*M^T product can be used for dimensionality reduction. This is because the resulting matrix is symmetric and positive semi-definite, and thus can be used as a covariance matrix. This covariance matrix can then be used to perform Principal Component Analysis (PCA), a common dimensionality reduction technique.

4. What are some applications of the M*M^T product in real-world problems?

The M*M^T product has many applications in real-world problems. Some examples include image processing, where it can be used for image compression and feature extraction; natural language processing, where it can be used for text classification and sentiment analysis; and machine learning, where it can be used for dimensionality reduction and feature selection.

5. Are there any limitations to using the M*M^T product of a matrix and its transpose?

One limitation of using the M*M^T product is that it can only be applied to square matrices. Additionally, it can be computationally expensive for large matrices, as it involves multiplying each row of the matrix by its own transpose. Furthermore, the resulting matrix can be ill-conditioned, meaning that small changes in the original matrix can result in large changes in the M*M^T product. Care should also be taken when using the M*M^T product for dimensionality reduction, as it may not always preserve the original data's most important features.

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