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If any n-dimensional nonzero vector is the eigenvector of n-order matrix A, then A is a quantitative matrix If any n-dimensional nonzero vector is the eigenvector of n-order matrix A, then A is a quantity matrix
It is proved that because any n-dimensional non-zero vector is the eigenvector of the n-order matrix A, the n-dimensional basic vector group ε1,ε2,...,εn is also the eigenvector of A. Let Aεi=kiεi, i=1,2,..., n then A (ε1,ε2,...,εn)=(Aε1, Aε2,..., Aεn)=(k1ε1, k2ε2,..., knεn)=(ε1,...
An example of what a matrix's row and column vector groups are For example, what are the row and column vector groups of a matrix
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Why b=λa, then a and b are collinear
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Vector a*b=/a/*/b/is collinear with vector a, b ---- condition
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For any vector b, vector a is collinear with b, then why is a
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If the vector a is parallel to the vector b, then the projection of the vector a on the vector b is the length of the vector a. On the other hand, if the vector PA+vector PB+vector PC=vector AB, then the position relation between point P and triangle ABC is (P is on the edge of AC). ? Get this problem solved, please! !! !! !! !! !! !! ! If the vector a is parallel to the vector b, then the projection of the vector a on the vector b is the length of the vector a. On the other hand, if the vector PA+vector PB+vector PC=vector AB, the position relation between point P and triangle ABC is (P is on the edge of AC). ? Get this problem solved, please! !! !! !! !! !! !! ! Vector a is parallel to vector b, then vector a is projected on vector b as the length of a, why not? On the other hand, if the vector PA+vector PB+vector PC=vector AB, then the position relation between point P and triangle ABC is (P is on the edge of AC). ? Get this problem solved, please! !! !! !! !! !! !! !
Vector a Parallel vector b indicates an included angle of 0° or 180°
The projection of vector a in vector b is |a|cos0° or |a|cos180°
I.e. the projection is a|
That is to say, there are two possibilities for equal lengths of projection!
Vector PA + Vector PB + Vector PC = Vector AB 1
Vector PB - Vector PA = Vector AB 2
Subtract two vectors PA+vector PC=0
Vector PA/vector PC=-1/2
Explain that p is at three points of AC and close to point A
Drawing is easy!
RELATED INFORMATIONS
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