increasing order≥A2≥.·≥Am.Hence,the projection functions of PCAH are defined as follows:f(x)=wix,where wi is the kth column of W. Let A =[A1,A2,...,Am]T and A diag(),where diag(A)denotes the diagonal matrix whose diagonal entries are formed from the vector A.It is easy to prove that WTXXTW =A.Hence,the variance of the values {f(x)1on the kth projected dimension,which corresponds to the kth row of WTX,is Ak.Obviously,the variances for different PCA dimensions are anisotropic. To get isotropic projection functions,the idea of our IsoHash method is to leam an orthogonal matrix Q E Rmxm which makes QTWTXXTWQ become a matrix with equal diagonal values, i.e..[QTWTXXTWQl =[QTWTXXTW]22 =..=[QTWTXXTWQlmm.Here.An denotes the ith diagonal entry of a square matrix A,and a matrix Q is said to be orthogonal if QQ=I where I is an identity matrix whose dimensionality depends on the context.The effect of the orthogonal matrix Q is to rotate the coordinate axes while keeping the Euclidean distances between any two points unchanged.It is easy to prove that the new projection functions of IsoHash are f(x)=(WQ)x which have the same (isotropic)variance.Here (WQ)k denotes the kth column of WQ. If we use tr(A)to denote the trace of a symmetric matrix A,we have the following Lemma 1. Lemma 1.If QTQ=I tr(QTAQ)=tr(A). Based on Lemma 1,we have tr(QTWTXXTWQ)=tr(WTXXTW)=tr(A)=>1Ai if QTQ I.Hence,to make QTWTXXTWQ become a matrix with equal diagonal values,we should set this diagonal value a= Let a=a1,2,,with4=a=∑ (1) m and T(z)={TE Rmxmldiag(T)=diag(z)}, where z is a vector of length m,diag(T)is overloaded to denote a diagonal matrix with the same diagonal entries of matrix T Based on our motivation of IsoHash,we can define the problem of IsoHash as follows: Problem 1.The problem of IsoHash is to find an orthogonal matrix Q making QTWTXXTWQE T(a),where a is defined in (1). Then,we have the following Theorem 1: Theorem 1.Assume QTQ I and T E T(a).If QTAQ =T,Q will be a solution to the problem of IsoHash. Proof.Because WTXXTW A,we have QTAQ =QT[WTXXTW]Q.It is obvious that Q will be a solution to the problem of IsoHash. As in [2],we define M(A)={QTAQJQ∈O(m)}. (2) where (m)is the set of all orthogonal matrices in Rmxm,i.e.,QTQ=I. According to Theorem 1,the problem of IsoHash is equivalent to finding an orthogonal matrix Q for the following equation [2]: IT-ZIF =0, (3) where T E T(a).Z E M(A).denotes the Frobenius norm.Please note that for ease of understanding,we use the same notations as those in [2. In the following content,we will use the Schur-Horn lemma [11]to prove that we can always find a solution to problem(3).increasing order λ1 ≥ λ2 ≥ · · · ≥ λm. Hence, the projection functions of PCAH are defined as follows: fk(x) = wT k x, where wk is the kth column of W. Let λ = [λ1, λ2, · · · , λm] T and Λ = diag(λ), where diag(λ) denotes the diagonal matrix whose diagonal entries are formed from the vector λ. It is easy to prove that WT XXT W = Λ. Hence, the variance of the values {fk(xi)} n i=1 on the kth projected dimension, which corresponds to the kth row of WT X, is λk. Obviously, the variances for different PCA dimensions are anisotropic. To get isotropic projection functions, the idea of our IsoHash method is to learn an orthogonal matrix Q ∈ R m×m which makes QT WT XXT W Q become a matrix with equal diagonal values, i.e., [QT WT XXT W Q]11 = [QT WT XXT W Q]22 = · · · = [QT WT XXT W Q]mm. Here, Aii denotes the ith diagonal entry of a square matrix A, and a matrix Q is said to be orthogonal if QT Q = I where I is an identity matrix whose dimensionality depends on the context. The effect of the orthogonal matrix Q is to rotate the coordinate axes while keeping the Euclidean distances between any two points unchanged. It is easy to prove that the new projection functions of IsoHash are fk(x) = (W Q) T k x which have the same (isotropic) variance. Here (W Q)k denotes the kth column of W Q. If we use tr(A) to denote the trace of a symmetric matrix A, we have the following Lemma 1. Lemma 1. If QT Q = I, tr(QT AQ) = tr(A). Based on Lemma 1, we have tr(QT WT XXT W Q) = tr(WT XXT W) = tr(Λ) = Pm i=1 λi if QT Q = I. Hence, to make QT WT XXT W Q become a matrix with equal diagonal values, we should set this diagonal value a = Pm i=1 λi m . Let a = [a1, a2, · · · , am] with ai = a = Pm i=1 λi m , (1) and T (z) = {T ∈ R m×m|diag(T) = diag(z)}, where z is a vector of length m, diag(T) is overloaded to denote a diagonal matrix with the same diagonal entries of matrix T. Based on our motivation of IsoHash, we can define the problem of IsoHash as follows: Problem 1. The problem of IsoHash is to find an orthogonal matrix Q making QT WT XXT W Q ∈ T (a), where a is defined in (1). Then, we have the following Theorem 1: Theorem 1. Assume QT Q = I and T ∈ T (a). If QTΛQ = T, Q will be a solution to the problem of IsoHash. Proof. Because WT XXT W = Λ, we have QTΛQ = QT [WT XXT W]Q. It is obvious that Q will be a solution to the problem of IsoHash. As in [2], we define M(Λ) = {Q TΛQ|Q ∈ O(m)}, (2) where O(m) is the set of all orthogonal matrices in R m×m, i.e., QT Q = I. According to Theorem 1, the problem of IsoHash is equivalent to finding an orthogonal matrix Q for the following equation [2]: ||T − Z||F = 0, (3) where T ∈ T (a), Z ∈ M(Λ), || · ||F denotes the Frobenius norm. Please note that for ease of understanding, we use the same notations as those in [2]. In the following content, we will use the Schur-Horn lemma [11] to prove that we can always find a solution to problem (3). 3